J + t = 30
j = 2t - 3
2t - 3 + t = 30
3t - 3 = 30
3t = 30 + 3
3t = 33
t = 33/3
t = 11 <=== tritt's age is 11
j = 2t - 3
j = 2(11) - 3
j = 22 - 3
j = 19 <== jan's age is 19
Answer : If it is perpendicular to the axis, then a circle. If it is at an angle to the axis, then an ellipse. If it is parallel to the axis, then two parallel lines. Those are the only 3 cases that I can think of.
Hope this helps.
9514 1404 393
Answer:
3 solutions: 1368, 5364, 9360
Step-by-step explanation:
In order to be divisible by 36, the number must be divisible by 9 and by 4. The number will be divisible by 9 if the sum of its digits is divisible by 9. 36 already has a digit sum of 9, so the two added digits must sum to 9.
The number will be divisible by 4 if the least-significant digit is a multiple of 4. That is, it must be 0, 4, or 8. Then the corresponding most-significant digits must be 9, 5, or 1.
The three solutions are ...
1368 = 38 · 36
5364 = 149 · 36
9360 = 260 · 36
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:
