Reflect the point across \(x = 5\).
I think your question is missing the numbers...
Here's another way:
The area of the 2 triangles is \(12 \times (5 + 4) \div 2 = 54 \)
The area of the white triangle is \(12 \times 5 \div 2 = 30\)
This means that the area of the shaded region is \(54 - 30 = \color{brown}\boxed{24}\)
Note that\((2a - 3)(4b - 6) = 8ab - 12b - 12a + 18 = 8ab - 12(a + b) + 18\)
By Vieta, \(a + b = {-6 \over 2} = -3\) and \(ab = {-14 \over 2} = -7\)
Plugging these in gives us \({8 \times -7 - 12(-3) + 18} = \color{brown}\boxed{-2}\)
What would make \(x^2 - 16\) negative?
What would make \(\sqrt{x^2 - 16} + 3\) negative?
Let the side of the cube be \(s\). We have the equation \(s^3 = 4 \times 12s\)
Solving the equation gives us \(s = \sqrt{48} = 4 \sqrt{3}\)
So, the volume of this is \((4 \sqrt 3)^3 = \color{brown}\boxed{192 \sqrt 3}\)
This simplifies to \(i^{14} = \color{brown}\boxed{-1}\)
The smallest vallue occurs at 100, or \(1 \times 15^2 + 0 \times 15 + 0 = \color{brown}\boxed{225}\)
Because the median is 20, we know that \(b = 20\).
Now, we want to minimize a, so we can maximize c.
Can you take it from here?
What diagram?
