Express each of the following pairs of signed numbers as absolute values and subtract the smaller absolute value from the larger absolute value. f.-33,7,-29.7

Prove directly from the definition that
${\displaystyle lim:\underset{x\to -2}{lim}{x}^2+x-5=-3}$

Simplify:
${\displaystyle \frac{-2+3x}{1+5x}}$
If ${\displaystyle x^2=-1}$

If a die is rolled 30 times, there are ${\displaystyle 6^{30}}$ different sequences possible.
What fraction of these sequences have exactly eight 1s and eight 6s?

${\displaystyle W_n=\frac{3n^{\sqrt{5}}-2n^3+n}{6n^{\sqrt{7}}+4n^3-8}}$, ${\displaystyle Y_n=\frac{6n^5+2n^2-1}{3n^2(2n^3+1)}}$, ${\displaystyle Z_n=1-\sin \left(\frac{2n\pi }{5}\right)}$

How to prove that if a quartic equation ( with real coefficients ) has 4 imaginary roots they all will be in conjugate pairs?
I proved this fact for qudratic equation in the following way , let a qudratic equation have a imaginary root ${\displaystyle p+iq}$(q is not 0), and let other root be ${\displaystyle (a+ib)}$. Now here sum of roots will be a real number lets say R, ${\displaystyle \Rightarrow (p+iq)(a-iq)=R^{\prime }\Rightarrow (pa+q^2)+i(aq-pq)=0\Rightarrow aq-pq=0\Rightarrow aq=pq\Rightarrow a=p}$. So finally the roots are, ${\displaystyle p+iq\text{and}p-iq}$, hence proved. I tried to prove this fact for quartic equation in a similar way but could not reach to the conclusion. Please guide me by answering by my method or by suggesting any other simple method to prove that if a quartic equation has all imaginary roots then they will occur in conjugate pairs.

Which of the sequences are bounded?
1) ${\displaystyle 4{(-1)}^n}$
2) ${\displaystyle 7n}$
3) ${\displaystyle \frac{1}{{13}^n}}$
4) ${\displaystyle \frac{n}{3n+1}}$

Adding and subtracting radicals
$\frac{\sqrt[3]{324a^4{b}^3}}{\sqrt[3]{6a{b}^2}}$