Rom

You'll need the input resistance of the TV

\(\text{The center of the original circle is }(0,0)\\ \text{The center of the reflected circle is }(8,2) \text{ (can you see why?)}\\ \text{The radius is unchanged. The equation of the new circle is }\\ (x-8)^2 + (y-2)^2 = 25\\ x^2 - 16x + 64 + y^2 - 4y + 4 = 25\\ x^2 +y^2 -16x-4y+43=0\\ a=1,~b=-16,~c=-4,~d=43\)

A = 2in x 3in = 6 in²

2+2+2+0+1 = 7

well done

\(x = n+1\\ \dfrac{dx}{dn}=1\\ y = 5+3n\\ \dfrac{dy}{dn}=3\\ \dfrac{dy}{dn} > \dfrac{dx}{dn}\\ y \text{ has the greater rate of change}\)

Another way you can do this is to note that

\(\text{For all }n\\ x_{n+1}-x_n = 1\\ y_{n+1}-y_n = 3\\ \text{and it's seen again that }y \text{ has the greater rate of change}\)

\(\text{Consider a triangle with legs }s_1=1,~s_2 = \sqrt{x^2-1}\\ \text{You end up with the equation}\\ \sec(\alpha) + \csc(\alpha) = \dfrac{35}{12}\\ \text{which can be solved with some annoying algebra}\)

\(1) \text{divide all terms by }2\\ 2) \text{consider a right triangle with legs }s_1 = \dfrac{\sqrt{3}}{2}x,~s_2 = \dfrac 1 2 x\)

\(\dfrac{df}{dx} = 0 \Rightarrow x = -2,~0\\ \text{we want } f(0) = -f(-2) = L\\ f(0) = N\\ f(-2) = N-16\\ N = 16-N\\ N=8 \)

\(f(x) \in \mathbb{R} - (-8,8)\)

\(f(6x) \text{ has the same range as }f(x)\\ f(6x)+1 \text{ shifts the range of }f(x) \text{ 1 unit in the positive direction }\\ \text{so that the range of }g(x) = f(6x)+1 \text{ is } [-10,4]\)