asiandude

The string can be divided into two different parts, the straight part and the curving part (when tangent to a circle). 

There are 6 straight parts, and each has a length of 2r. The total length is 2r * 6 = 12r.

The curving part adds up to the length of one of the circles, so its length should be $\pi r^2$.

Together the length is $12r+\pi r^2$.

If r is incremented by 2, we just have to replace r with (r+2). The length after incrementing is $12(r+2)+\pi(r+2)^2$.

Simplify, the length is $\pi r^2+12r+4 \pi r +28$.

The difference between the two lengths are $(\pi r^2+12r+4 \pi r +28) - (\pi r^2+12r)=4 \pi r+28$. And the string would be that much longer. 

We can call two adjacent terms a pair. For example, 2012 and -2002 is a pair and 32, -22 is also a pair. And the sum of the two values in each pair is 10. There are 101 pairs in totally and the sum of all the pairs should be$101\cdot10=1010.$

That's a circle. 

CXZ forms a triangle. And two sides of a triangle (XZ and ZC) will never be parallel, so this question is probably incorrect. 

It's a hyperbola.

Ok, the sum will equal 9 in that case. But I think there are other cases too, for example when x^2-4x+2 = 1. Can you help me think of other ones? 

sorry, it's $(x^2-4x+2)^{x^2-5x+2}=1$

$\frac{\sqrt2+\sqrt3}{\sqrt{10}+\sqrt{14}+\sqrt{15}+\sqrt{21}}$

Looking at the denominator, notice that $\sqrt{10}$ and $\sqrt{14}$ are both divisible by $\sqrt{2}$ and$\sqrt{15}$, $\sqrt{21}$ are divisible by $\sqrt{2}$. 

We can factor those out to get $\frac{\sqrt2+\sqrt3}{\sqrt2(\sqrt5+\sqrt7)+\sqrt{3}(\sqrt5+\sqrt7)} = \frac{\sqrt2+\sqrt3}{(\sqrt2+\sqrt{3})(\sqrt5+\sqrt7)}$. Divide top and bottom by $\sqrt2+\sqrt3$: $\frac{1}{\sqrt5+\sqrt7}$.

Moving the radicals to the numerator: $\frac{1}{\sqrt5+\sqrt7} = \frac{1\cdot(\sqrt5-\sqrt7)}{(\sqrt5+\sqrt7)(\sqrt5-\sqrt7)} = \frac{\sqrt{7}-\sqrt{5}}{2}$.

So the final result is: $\boxed{\frac{\sqrt7 - \sqrt5}{2}}$. Hope this helped 

The equation would be y = 2x-16. 

So when x = 0, y = -16.

(0, -16)