thedudemanguyperson

At $x=-8.4$ in the second equation, $y=14+4.2=18.2$. Then plug this into the first equation along with $x$ to find $k$.

$62910=351261_7=172676_8$

Take the entire thing mod $10$.

It's then just $9(1^{2009}+...+9^{2009})\pmod{10}$.

And $1^{2009}=(-9)^{2009}=-9^{2009}\pmod{10}$.

Similarly, $2^{2009}=-8^{2009}\pmod{10}$, etc.

So everything cancels and it becomes $9\cdot 5^{2009}\pmod{10}$. Note that $5^n$ is always $5$ modulo $10$. Then the answer is $9\cdot 5=5\pmod{10}$.

(mod 10 is just the remainder when divided by 10)

Here's a more braindead but easy solution with mass points. Assign $A=2$ and $C=1$. Then $D=3$, and $B=3$ so $E=6$ then $F=4$. Hence $[AFC]=\frac{800}{3}$, and then $[BEF]=400\cdot 1/3\cdot 1/2=\frac{200}{3}$. So $[EBA]=400-\frac{800}{3}-\frac{200}{3}=\boxed{\frac{200}{3}}$.

I assume (?) you want to eat as many cheez-it's as possible.

In that case, the width of the first is $\frac{22}{2}-9=2$, and the width of the second is $\frac{20}{2}-5=5$.

So it's a $9\times 2$ rectangle and a $5\times 5$ rectangle. Which has greater area? (In other words, which of $9\times 2$ and $5\times 5$ is larger)

The height could be any positive number that you want it to be

AMSP huh?

Upon division it's simply $3x-1$. The only value that this can't be is when $x=1$, and that is $3(1)-1=\boxed{2}$.

A) Just plug in the formula

B) What interval?

Let $OA=OB=OC=x$. Then $x^2\sin\theta=100$ by sine area formula (I used $\angle AOB=\theta$). By Law of Cosines, $2x^2-2x^2\cos\theta=100\implies x^2(1-\cos\theta)=50$. So dividing, $\frac{1-\cos\theta}{\sin\theta}=\frac{1}{2}\implies \tan(\theta/2)=1/2$. So it's not hard to get $\sin\theta=\frac{4}{5}$ (it should remind you of the 3-4-5 triangle). Then $x^2=100*5/4=125\implies x=\sqrt{125}=\boxed{5\sqrt{5}}$

The other answer is utter nonsense.