Vinculum

Sorry!

Its more complicated than I thought!!!

First, its good to know that triangle PRQ and MNP are similar 

Angle PQR=RPQ are 58deg

Since PRQ and MNP are similar, both angles RPQ and NPM are 58deg

Since this is an isosceles triangle...

58deg+2(anglePMN)=180

So we get...

2(anglePMN) = 122

Angle PMN = 61 degrees

\(8+12=20\)

So the sum of the lengths of the two longest altitudes in that triangle is \(20\)

Kinda sus man!!!!

Hopefully, this will help:

https://web2.0calc.com/questions/counting_31398

The volume of a cone is 1/3hπr²

The height is 15 and the radius must be 7.5.

Substituting these values in the formula gives 

1/3*15*π*(7.5)^2

This gives approx. 883.57

\(\left|x-5\right|+11\le \:18\)

\(\left|x-5\right|+11-11\le \:18-11\)

\(\left|x-5\right|\le \:7\)

\(\mathrm{Apply\:absolute\:rule}:\quad \mathrm{If}\:|u|\:\le \:a,\:a\:>\:0\:\mathrm{then}\:-a\:\le \:u\:\le \:a\)

\(-7\le \:x-5\le \:7\)

\(x\ge \:-2\quad \mathrm{and}\quad \:x\le \:12\)

\(== -2\le \:x\le \:12\)

Area of cylinder: \(2\pi r^2+2\pi rh\)

\(2\pi \:\:r^2+2\pi \:\:r\left(2.5r\right)\)

\(7\pi \:r^2=448\pi \)

\(r=-8,\:r=8\)

But a radius can't be negative so our radius is \(8\)!!!

\(g(x) = \sqrt{16+9}=5\)

Sub \(5\) into function \(t(x)\):

\(3-\sqrt{5+9}\) \(=3-\sqrt{14}\)

Answer is \(3-\sqrt{14}\)

you beat me to it CPhill