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3. Let's try by completing the square. We simplify the original expression, to $x^2-12x=-20$. Then, solve for $a$ while completing the square, so $2ax=-12x, a=-6$ . Finally, you end up with $\left(x-6\right)^2=16$, therefore the only solutions are $\boxed{x=2, x=10}$ .

1. First, try to change the x in the middle, to a fraction. Let's say x/y. We then multiply both sides by b, to get 9b/11 and 11b/13. Now, the trick here is to find a positive integer that satisfies and is less than 1, since 11/13 is less than 1. 

Hello! Here is a long way!

If you know the principle of Inclusion and Exclusion, this will lead to the size of their union, but in this case, we have to find the intersection! Therefore, we can find the size of union and subtract from the number of multiples of 4 and 6. There are $\frac{170}{4}\approx42$ multiples of $4$ and $\frac{170}{6}\approx28$ multiples of $6.$The least common multiple of 4 and 6 is 12, so $\frac{170}{12}\approx14$ . You could stop here, because this is your answer, but let's continue and show why this works. The union is just $42+28-14=56$ and there are $42+28$ total multiple of 4 and 6. Thus, the intersection is $70-56=\boxed{14}$ . You could've stopped well before and found your answer, but here is just a long way. 

Good job, EP! I'll just go over the formula. These steps are called PRINCIPLE OF INCLUSION AND EXCLUSION, and it comes up in many problems. Just what EP said, overcounts are there, and you have to remove them!

We can break this up into: $(130+3)(130+3)$ . Now, by distributing, we have $16900+390+390+9=\boxed{17689}$ .

(b) We set up an equation: $\frac{2}{3}x=14 \longrightarrow x=14*\frac{3}{2} \longrightarrow \boxed{x=21}$ games.

Since we know the volume of a cone, we just plug in our values! The radius is 8cm and the vertical height is 15cm. Thus, $\frac{1}{3}\pi*64*15=\frac{1}{3}*\pi*960=1004.8$

This is the same as $\frac{100!}{2!*98!}=\frac{100*99}{2*1}=50*99=\boxed{4950}.$

We know the midpoint formula is $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$ . Plugging our values in, we get $(\frac{4+x_2}{2}, \frac{5+y_2}{2})=(1,-2)$ . Solving for $x_2,$ we attain $\frac{4+x_2}{2}=1, 4+x_2=2, x_2=-2$ and $\frac{5+y_2}{2}=-2, 5+y_2=-4, y_2=-9$ . Thus, the order pair is $\boxed{(-2,-9)}$ .

Thank you, CPhill, for providing a wonderful solution! I was actually stumped at first..