$$\frac{2.10^{-7} - 0,4.10^{-6}}{10^{-8}} = ? $$
These questions are making me confused because we're dealing with the terms like $10^x$. What are your professional tips?
My attempt:
$$\frac{2.10^{-7} - 4.10^{-7}}{10^{-8}} \tag{1} $$ $$\frac{ -8.10^{-7}}{10^{-8}} \tag{2} $$
And that's where I'm stuck.
$(2)$ is incorrect.
From $(1)$, $\displaystyle \frac{2\cdot10^{-7} - 4\cdot10^{-7}}{10^{-8}}=\frac{(2-4)\cdot 10^{-7}}{10^{-8}}=\frac{-2\cdot 10^{-7}\cdot 10^8}{10^{-8}\cdot 10^8}=\frac{-2\cdot 10}{1}=-20$
From here
$$\frac{2\cdot 10^{-7} - 4\cdot 10^{-7}}{10^{-8}}=\frac{-2\cdot 10^{-7}}{10^{-8}}=-2\cdot 10^{-7}\cdot 10^{8}=-20$$
or as an alterntive
$$\frac{2\cdot 10^{-7} - 0.4\cdot 10^{-6}}{10^{-8}}=2\cdot 10^{-7}\cdot 10^{8} - 0.4\cdot 10^{-6}\cdot 10^{8}=20-40=-20$$
multplying numerator and denominator by $10^{8}$ we get $$2\cdot 10^{-7}\cdot 10^{8}-0.4\cdot 10^{-6}\cdot 10^{8}$$ and we get $$2\cdot 10-0.4\cdot 10^2=...$$
$$\frac{2.10^{-7} - 0,4.10^{-6}}{10^{-8}} = 10^8 (2.10^{-7} - 0,4.10^{-6})=20-40=-20$$