In a right triangle $ABC$, $tan(A)=\frac{5}{12}$. Find $sin(A)$.

5 12 13 Let $a$ be the length of the side opposite of angle $A$, $b$ be the length of the side adjacent to angle $A$, and $h$ be the length of the hypotenuse. We know that $tan(A)=\frac{a}{b}$ and therefore $a=5k$ and $b=12k$ where $k$ is a coefficient. Since we know these two side lengths, we can find the length of the hypotenuse using the Pythagorean Theorem. $h^2=(5k)^2+(12k)^2$ $h^2=(25+144)k^2$ $h^2=169k^2$ $h=13k$ Since $sin(A)=\frac{a}{h}$, we know that $sin(A)=\frac{5k}{13k}=\frac{5}{13}$

Solve for $x$. $52*x\equiv 121 (mod 11)$

Since $121\equiv0 (mod 11)$ we can rewrite the equation as $52*x\equiv 0 (mod 11)$. Dividing by 52 on both sides we find that $x\equiv 0 (mod 11)$.

Use L'Hopital's Rule to evaluate the following limit. $\lim_{x \to \infty} \frac{x}{x^2+1}$

Since $\lim_{x \to \infty} \frac{x}{x^2+1}$ yields $\frac{\infty}{\infty}$ which is undefined, we must use L'Hopital's Rule to determine the limit. Taking the derivative of the numerator and the denominator of the function yields a new limit: $\lim_{x \to \infty} \frac{1}{2x+1}=\frac{1}{\infty}=0$. Therefore $\lim_{x \to \infty} \frac{x}{x^2+1}=0$