Part A:
Given that
![0.8x=1.6y\Rightarrow x= \frac{1.6}{0.8} y=2y](https://tex.z-dn.net/?f=0.8x%3D1.6y%5CRightarrow%20x%3D%20%5Cfrac%7B1.6%7D%7B0.8%7D%20y%3D2y)
Thus, x = ky where k = 2.
Therefore, the given equation is a direct variation.
Part B:
If y varies directly as x, then y = kx.
Given that y = 5 when x = 2, then
![5=2k \\ \\ \Rightarrow k=\frac{5}{2} =2.5 \\ \\ \Rightarrow y=2.5x](https://tex.z-dn.net/?f=5%3D2k%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20k%3D%5Cfrac%7B5%7D%7B2%7D%20%3D2.5%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20y%3D2.5x)
When x = 12, y = 2.5(12) = 30.
Part C:
If y varies directly as x, then y = kx.
Given that y = 9 when x = -6, then
![9=-6k \\ \\ \Rightarrow k=-\frac{9}{6} =-\frac{3}{2} \\ \\ \Rightarrow y=-\frac{3}{2}x](https://tex.z-dn.net/?f=9%3D-6k%20%5C%5C%20%5C%5C%20%5CRightarrow%20k%3D-%5Cfrac%7B9%7D%7B6%7D%20%3D-%5Cfrac%7B3%7D%7B2%7D%20%5C%5C%20%5C%5C%20%5CRightarrow%20y%3D-%5Cfrac%7B3%7D%7B2%7Dx)
It can be seen that the equation above satisfies all the rows of the given table, therefore, y varies with x for the data in the question and the equation for the direct variation is
![y=-\frac{3}{2}x](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B3%7D%7B2%7Dx)
Part 4:
If y varies directly as x, then y = kx.
Given that y = 1 when x = -2, then
![1=-2k \\ \\ \Rightarrow k=-\frac{1}{2} \\ \\ \Rightarrow y=-\frac{1}{2}x](https://tex.z-dn.net/?f=1%3D-2k%20%5C%5C%20%5C%5C%20%5CRightarrow%20k%3D-%5Cfrac%7B1%7D%7B2%7D%20%5C%5C%20%5C%5C%20%5CRightarrow%20y%3D-%5Cfrac%7B1%7D%7B2%7Dx)
It can be seen that the equation above does not satisfies the other rows of the given table, therefore, y does not vary with x for the data in the question.