thedudemanguyperson

Hint: Radius $r$ of the circle, side $s$ of the square implies that $2r=s$.

$x^2+2x+y^2-11y+8=(x+1)^2+\left(y-\frac{11}{2}\right)^2+8-1-\frac{11^2}{2^2}\ge 8-1-\frac{121}{4}=-\frac{93}{4}$. 

Notice that $2013\equiv 11\equiv -2\pmod{13}$.

So $2013^{2013^2}\equiv (-2)^{2013^2}\equiv -2^{2013^2}\pmod{13}$ since $2013^2$ is odd.

Notice that $\gcd(2,13)=1$ and $13$ is prime, so we can invoke fermat's little theorem, i.e. $2^{12}\equiv 1\pmod{13}$.

Notice that $2013\equiv 9\pmod{12}\implies 2013^2\equiv 9^2=81\equiv 9\pmod{12}$.

Hence $-2^{2013^2}\equiv -2^9\pmod{13}$. Note $2^9\equiv 512\equiv 5\pmod{13}$, hence the answer is $-5\equiv \boxed{8}\pmod{13}$.

From the rational root theorem, the denominator of the root divides 27 and the numerator divides 9. Keeping in mind that negatives are allowed, the answer is $(2d(9))(2d(27))$ where $d(x)$ denotes the number of divisors of $x$ (which is really easy to manually calculate).

Your truck: $T$

Neighbor: $t$

$T\ge t+600$

$T\le \frac{4t}{3}$

Solve the inequality chain.

Write everything in terms of powers of $2$. Then use that $2^n<2^N$ if and only if $n

$\frac{1}{6}$ for first and $\frac{1}{6}$ for second. Multiply the probabilities.

The parallel condition implies that it has slope $-\frac{1}{3}$. What can you determine about $A$ and $B$ given this? Now substitute $-7A+2B=3$ with your first equation to explicitly find $A,B$ and in turn find $B-A$.

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc=3abc=54$ since the first term is negligible as $a+b+c=0$.

cleannnn