multiply x times y divide by 6 on both sides and you got your answer
The equation of direct variation is,

Substitute the values of <em>y</em> and <em>x</em>.

So, the obtained equation will be <em>y</em>=1.35(x).
Let's check with the values of <em>x</em>.
1) If <em>x</em>=4

2) If <em>x</em>=2

3) If <em>x</em>=-2
Answer:
The answer is 
Step-by-step explanation:
To calculate the volumen of the solid we solve the next double integral:

Solving:

![[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.](https://tex.z-dn.net/?f=%5B6x%5E%7B2%7D%20%5D%7B%7B1%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.%20%2A%20%5B%5Cfrac%7By%5E%7B3%7D%7D%7B3%7D%5D%7B%7B1%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
Replacing the limits:

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.
Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ 
Solving the double integral with these new limits we have:

This part is a little bit tricky so let's solve the integral first for dy:
![\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%5Cfrac%7B1%7D%7Bm%7D_0%20%5B%7B12x%20%5Cfrac%7By%5E%7B3%7D%7D%7B3%7D%7D%5D%7B%7B1%7D%20%5Catop%20%7Bmx%7D%7D%20%5Cright.%5C%2C%20dx%20%3D%5Cint%5Climits%5E%5Cfrac%7B1%7D%7Bm%7D_0%20%5B%7B4x%20y%5E%7B3%20%7D%5D%7B%7B1%7D%20%5Catop%20%7Bmx%7D%7D%20%5Cright.%5C%2C%20dx)
Replacing the limits:

Solving now for dx:
![[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.](https://tex.z-dn.net/?f=%5B%7B%5Cfrac%7B4x%5E%7B2%7D%7D%7B2%7D%20-%5Cfrac%7B4m%5E%7B3%7D%20x%5E%7B5%7D%7D%7B5%7D%20%5D%7B%7B%5Cfrac%7B1%7D%7Bm%7D%20%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.%20%3D%20%5B%7B2x%5E%7B2%7D%20-%5Cfrac%7B4m%5E%7B3%7D%20x%5E%7B5%7D%7D%7B5%7D%20%5D%7B%7B%5Cfrac%7B1%7D%7Bm%7D%20%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
Replacing the limits:

As I mentioned before, this volume is equal to 1, hence:

20*6=120seconds +3600second in an hour so 3600*4=14400
1210+14400=14520
14520 is the answer I hope this helps!
Slope:7
y-int:-2
slope is the one with x and y-int is b