A quadrilateral QNML is a trapezoid, therefore |∡QNM| + |∡NML| = 180°
|∡NML| = 180° - |∡QNM| ⇒ |∡NML| = 180° - 82° = 98°
|∡NML| = |∡NMP| + |∡PML| ⇒ |∡NMP| = |∡NML| - |∡PML|
|∡NMP| = 98° - 30° = 68°
Answer: |∡NMP| = 68° .
I hope this helps you! Feel free to ask ant questions
Answer:
the radius of the base is equal to ![x=2\sqrt[3]{\frac{105}{\pi } }](https://tex.z-dn.net/?f=x%3D2%5Csqrt%5B3%5D%7B%5Cfrac%7B105%7D%7B%5Cpi%20%7D%20%7D)
And the height is equal to:
![y=\frac{840}{\pi(2\sqrt[3]{\frac{105}{\pi}})^{2} }](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B840%7D%7B%5Cpi%282%5Csqrt%5B3%5D%7B%5Cfrac%7B105%7D%7B%5Cpi%7D%7D%29%5E%7B2%7D%20%7D)
Step-by-step explanation:
We write the volume function
where x is the radius of the base and y is the height of the cylinder

The surface of a cylinder is given by
on the interval from 0 to infinity
We now determine the critical values by differentiating and making the equation equal to 0

This equation have 2 complex solutions and one real solution
![x=2\sqrt[3]{\frac{105}{\pi } }](https://tex.z-dn.net/?f=x%3D2%5Csqrt%5B3%5D%7B%5Cfrac%7B105%7D%7B%5Cpi%20%7D%20%7D)
For x=0 and infinity the equation goes to infinity also so the radius of the base is equal to ![x=2\sqrt[3]{\frac{105}{\pi } }](https://tex.z-dn.net/?f=x%3D2%5Csqrt%5B3%5D%7B%5Cfrac%7B105%7D%7B%5Cpi%20%7D%20%7D)
And the height is equal to:
![y=\frac{840}{\pi (2\sqrt[3]{\frac{105}{\pi } })^{2} }](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B840%7D%7B%5Cpi%20%282%5Csqrt%5B3%5D%7B%5Cfrac%7B105%7D%7B%5Cpi%20%7D%20%7D%29%5E%7B2%7D%20%7D)
![y=\frac{840}{\pi(2\sqrt[3]{\frac{105}{\pi}})^{2} }](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B840%7D%7B%5Cpi%282%5Csqrt%5B3%5D%7B%5Cfrac%7B105%7D%7B%5Cpi%7D%7D%29%5E%7B2%7D%20%7D)
Answer:
12
Step-by-step explanation:
4,8,12
6,12
They both have 12 in common
The answer is D. 60'. You add both angles together, then subtract them from 180.
60+70=120
180-120=60