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Variation constants and exact value
Talk about a nightmare! I just finally understand one thing, and this comes along. The first problem I finished easily because all of the numbers were supplied, but the second problem has me stumped. The original problem reads like this,
Let y=ax^1/4. The first problem was y=1.2 and x =16. That one was pretty easy but the second is find the exact value for y if x=6. I know that you plug in 6 for x, but I guess I am confused as to what exact value and direct variation and variation constants are. I am not sure how to set the second problem up?? And then you have to find x if y=1.8 I think that I would know how to set them up if I knew exactly what they were, but the textbook does not give a good explanation, nor did my professor. :confused: |
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