CPhill

Nice answer, Stormegetton......good humor, too !!!

-18+4+x=9-12    simplify

-14 + x = -3        add14 to both sides

x = 11

(1 + cx) (1 + x)^4

Expanding (1+ x)^4 we have

(1 + 4x + 6x^2 + 4x^3 + x^4)

So  (1 + cx)(1 + 4x + 6x^2 + 4x^3 + x^4) produces two x^3 terms, i.e. 4x^3 and 6cx^3

So

4 + 6c = 22     subtract 4 from both sides

6c = 18           divide both sides by 6

c = 3

Melody, I only know how to do (a) using L'Hospital's   

Taking the derivative of the top and the bottom =

(1/2)(2)/√(2x + 4) = 1/√(2x + 4)

So

lim x → 0   1/√(2x + 4)  = 1/2

For (c)....note that     -1 ≤ sin x^2 ≤ 1    ....and I believe we can do this one like (e) using the Squeeze Theorem

d.n.e. = does not exist

97a^3+0.564a=0  ....  factoring, we have

a(97a^2 + .564) = 0     the only real solution is a = 0

Note substituting x for a and using the on-site solver, we have

97x^2 + .564 = 0

$${\mathtt{97}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.564}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{141}}}}{\mathtt{\,\times\,}}{i}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{970}}}}\right)}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\mathtt{141}}}}{\mathtt{\,\times\,}}{i}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{970}}}}\right)}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.076\: \!252\: \!429\: \!401\: \!893\: \!3}}{i}\\
{\mathtt{x}} = {\mathtt{0.076\: \!252\: \!429\: \!401\: \!893\: \!3}}{i}\\
\end{array} \right\}$$

9 = 3²

Let's simplify -1+i/i+1   multiply numerator and denominator by i-1...so we have

[i - 1][i-1]/[(i+1)(i-1)] = (i^2 - 2i + 1)/(i^2 -1) = (-2i) / (-2) = i =  0 + 1i

tan(1/0) = undefined = pi/2

So...arg(z) = pi/2 + n*(2pi) for n = 0, ±1,±2, ±3.......

560/100=a^17

We can write this as

28/5 = a^17       take the log of both sides

log(28/5) = log a^17    and we can write

log(28/5) = 17 log a     divide by 17 on both sides

log(28/5)/17 = log a     and exponentially, we have

10^log(28/5)/17 = a = about 1.10665

2x^2-12x+13

This polynomial is irreducible into linear factors......thus it is not "factorable"