anonymous
  • anonymous
suppose y varies directly with x. if y=-3 when y=5,find y when x=-1
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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dan815
  • dan815
i think something wrong in ur question
dan815
  • dan815
suppose y varies directly with x. if y=-3 when X*?=5,find y when x=-1
anonymous
  • anonymous
I think that the answer is y=3

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anonymous
  • anonymous
x y -1 3 0 2 1 1 2 0 3 -1 4 -2 5 -3
whpalmer4
  • whpalmer4
If y varies directly with x, then \(y = kx\), \(k\) some constant. \(-3 = k *5\rightarrow k=-\frac{3}{5}\) \(y = k(-1)\)\[y = -\frac{3}{5}(-1) = \frac{3}{5}\] @cahayes9498 with direct variation, you've just got a straight line, and the slope is \(k\). Here \(k = -\frac{3}{5}\) so the line slopes down and to the right. Your table would be correct if we knew that \(k = -1\), but it doesn't, so the relationship between x and y is not correct.
anonymous
  • anonymous
@whpalmer4 you have me completely lost now, where is the K coming in at?
whpalmer4
  • whpalmer4
Any variation problem has a scale factor in there. We have to find it using the initial data. So, because it is direct variation, we know the form is \(y = kx\). If we double \(x\), \(y\) also doubles. That's direct variation. Now, the reason the \(k\) factor is there is so we can have direct variation between any possible pair of \(x\) and \(y\) values. Say I had two quantities that exhibited direct variation, but when the \(x\) value was 1, the \(y\) value was 3. How do you make that happen, except by a scale factor? You can't just add (which is what your table did) because then doubling \(x\) won't double \(y\).
whpalmer4
  • whpalmer4
Looking at it another way: with this problem, we knew that y = -3 when x = 5. Well, if it is direct variation, then doubling \(x\) must also double \(y\). If we evaluate my function at x = 10, we get \(y = -\frac{3}{5}(10) = -6\) which is twice the value it is at x = 5. With your table, y would = -8, and that isn't a doubling of what it was at x = 5.
whpalmer4
  • whpalmer4
Does that make sense?
anonymous
  • anonymous
ok I understand it now, yes it makes sense now
whpalmer4
  • whpalmer4
Good! Are you comfortable with the difference between direct, indirect and joint variation?
anonymous
  • anonymous
yes
whpalmer4
  • whpalmer4
Also good. Each of them will have that pesky \(k\) in the numerator.
whpalmer4
  • whpalmer4
Have you taken any physics classes?
whpalmer4
  • whpalmer4
Lots of variation examples there, and although the variation constant isn't usually called \(k\), it's always there. Perhaps the most obvious one is Newton's F=ma — for a given mass, the acceleration a and force F vary directly, and m serves as the variation constant.
anonymous
  • anonymous
no I have not
whpalmer4
  • whpalmer4
Another example might be the distance formula: distance = speed * time. If you go at a steady speed, the distance varies directly with the time. Double the time, double the distance. Here the speed is the variation constant. I think I've beaten this horse to death, or at least bored it :-)

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