\(\text{RECTANGLE}\\ \begin{align} A&=\sqrt{5}\\ a&=1+\sqrt{5}\\ A&=ab\\ b&=\frac{A}{a}\\ b&=\frac{\sqrt{5}}{1+\sqrt{5}}\\ b & \approx 0.690983\\\\ \text{answer} &= a+b \sqrt{5}\\ \text{answer} &= 1\sqrt{5}+0.690983 \sqrt{5}\\ \text{answer} &=4.781153 \end{align}\)
\(\begin{align} 5+5-\frac{5^2}{5 \cdot 36}&=\\ 5+5-\frac{25}{180}&=\\ 5+5-0,13\bar{8}&\approx\\ 5+5-0,139&=\\ 10-0,139&=\\ 9,861 \end{align}\)
The simplest way to write down your mathematica statement is by using factorials:
\(1 \cdot 2 \cdot 3 \cdot 4 \cdot 5=5!=120\)
I would say 2.
The thing that confuses me is your "opinion" on what can be regarded as a point. There are two limits in that function thta I didn't include, because they are technically undefinable.
Here's the graph: https://www.desmos.com/calculator
Because:
\(\begin{align} 4-5x(-1)&=9\\ 4+5x&=9\\ 5x&=9-4\\ x&=\frac{5}{5}\\ x&=1 \end{align}\)
You do it like this:
\(\begin{align} \frac{3122}{1}\cdot \frac{3}{100}&=\\ \frac{3122\cdot 3}{100}&=\\ 93.66 \end{align}\)
Like this:\(\begin{align} -\frac{x}{2}-4&=-7 \; /\cdot(-2)\\ \frac{x(-2)}{-2}-4(-2)&=-7(-2)\\ x+8&=14\\ x&=14-8\\ x&=6 \end{align}\)
