thedudemanguyperson

\(x-3\ne 0\rightarrow x\ne 3\). Also, \(2x-5\ge 0\rightarrow x\ge 5/2\). So domain is \(x\ge 5/2, x\ne 3\). For range, it is simply all reals.

First of all, x-7/=0 --> x/=7. Also, 7(x/(x-7))/=6 --> 7x/=6x-42 --> x/=-42. So \(\{x|x\in\mathbb{R}, x\ne 7, x\ne -42\}\)

slope is (9-7)/(-4-3)=-2/7. So y=-2/7x+b --> 7=-2/7*3+b --> b=55/7. So y=-2/7x+55/7 --> when y=1 --> 1=-2/7a+55/7 --> a=24

\(8\sqrt2-\sqrt{24}-\sqrt3+\sqrt{128}=8\sqrt2-2\sqrt6-\sqrt3+8\sqrt{2}=16\sqrt{2} - \sqrt{3} - 2 \sqrt{6}\)

Nice. 

\(\frac{a}{1-r}=3 \rightarrow a=3-3r\) by geometric sum formula, where r is the common ratio. Our sequence would be \(a,ar,ar^2,ar^3,...\). When squared, it would be \(a^2, a^2r^2, a^2r^4, a^2r^6\). This sequence has first term a^2 and common ratio r^2, so by the geometric sum formula this is \(\frac{a^2}{1-r^2}=3 \rightarrow \frac{(3-3r)^2}{1-r^2}=3\rightarrow 9r^2-12r+9=3-3r^2 \rightarrow r=\frac{1\pm i}{2}\). So also \(a=3-\frac{3\pm 3i}{2}=\frac{3\pm 3i}{2}\).

Let z=x^2 --> z^2-5z-36=0 --> (z+4)(z-9)=0 --> z=-4,9 --> x^2=-4 (no real solutions), x^2=9 --> x=3,-3

If you wanted imaginary solutions as well --> x^2=-4 --> x=2i,-2i --> x=3,-3,2i,-2i

\(2\pi r*h=2\pi *5*2=20\pi\). Now for top and bottom. Top is \(5^2\pi-2^2\pi=25\pi-4\pi=21\pi\), and so is bottom. So ans is \(20\pi+21\pi+21\pi=\boxed{62\pi}\).

I don't like point-slope so I will convert.

3x-2y=-9 --> 2y=3x+9 --> y=3/2x+9/2

Perpendicular has slope -2/3. So y=-2/3x+b. Plug (-4,-2) --> -2=(-2/3)*(-4)+b --> -2=8/3+b --> b=-14/3

So eqn. is y=-2/3x-14/3 --> 3y=-2x-14 --> 3y+2x+14=0

Case one:

4(x+7)+3x+2=32 --> 4x+28+3x+2=32 --> 7x=2 --> x=7/2

Case two:

-4(x+7)+3x+2=32 --> -4x-28+3x+2=32 --> -x=58 --> x=-58

x=7/2, -58

Note that \((x+y)^2-2xy=100-2xy=x^2+y^2=56\rightarrow 100-2xy=56\rightarrow xy=22, x+y=10\). So x and y are the roots to this quadratic: \(z^2-10z+22=0\rightarrow z^2-10z+25=3\rightarrow (z-5)^2=3\rightarrow z=5\pm \sqrt{3}\rightarrow\boxed{(x,y)=(5+\sqrt3,5-\sqrt3), (5-\sqrt3, 5+\sqrt3)}\).