Answers:
- Inverse variation
- Direct variation
- Direct variation
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Step-by-step explanation:
Direct variation equations follow the format of y = kx, where k is some constant. Isolating k gets us k = y/x. The x value cannot be zero.
The idea is that we divide each pair of (x,y) coordinates. If we get the same result each time, then we have confirmed the equation is a direct variation.
Inverse variation equations are of the form y = k/x in which we can solve for k to get k = xy. This time we multiply the x and y values. If we get the same thing each time then we have an inverse variation equation.
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Let's look at the table in problem 1.
Dividing the x and y values produces the following
- Row One: y/x = 10/2 = 5
- Row Two: y/x = 5/4 = 1.25
We don't get the same result, so table1 is NOT a direct variation.
What about an inverse variation? Well we multiply the x,y values this time
- Row One: xy = 2*10 = 20
- Row Two: xy = 4*5 = 20
- Row Three: xy = 5*4 = 20
- Row Four: xy = 20*1 = 20
Luckily we get the same result each time, so this confirms table1 is an inverse variation.
The equation for table1 is y = 20/x
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Now onto problem 2.
Let's check if we have a direct variation.
Divide the y over x values
- Row One: y/x = 6/1 = 6
- Row Two: y/x = 12/2 = 6
- Row Three: y/x = 30/5 = 6
- Row Four: y/x = 42/7 = 6
We get the same result each time. We found that k = 6, so the direct variation equation is y = 6x. Whatever x is, multiply by 6 to find y.
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Lastly problem 3
We'll see if we have a direct variation.
Once again, divide like we did in the previous problem
- Row One: y/x = 25/0.2 = 125
- Row Two: y/x = 62.5/0.5 = 125
- Row Three: y/x = 250/2 = 125
- Row Four: y/x = 375/3 = 125
Therefore, k = y/x = 125 and we go from y = kx to y = 125x
This is a direct variation equation.
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Summary:
Problem 1 is an inverse variation
Problems 2 and 3 are direct variation