Question

# Which statement is correct? (3.56*10^2)/(1.09*10^4)<=(4.08*10^2)(1.95*10^-6) (3.56*10^2)/(1.09*10^4)<(4.08*10^2)(1.95*10^-6) (3.56*10^2)/(1.09*10^4)>(4.08*10^2)(1.95*10^-6) (3.56*10^2)/(1.09*10^4)=(4.08*10^2)(1.95*10^-6)

Complex numbers
Which statement is correct?
$$\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}\le{\left({4.08}\cdot{10}^{{2}}\right)}{\left({1.95}\cdot{10}^{{-{{6}}}}\right)}$$
$$\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}{<}{\left({4.08}\cdot{10}^{{2}}\right)}{\left({1.95}\cdot{10}^{{-{{6}}}}\right)}$$</span>
$$\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}{>}{\left({4.08}\cdot{10}^{{2}}\right)}{\left({1.95}\cdot{10}^{{-{{6}}}}\right)}$$
$$\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}={\left({4.08}\cdot{10}^{{2}}\right)}{\left({1.95}\cdot{10}^{{-{{6}}}}\right)}$$

2020-11-01
Let us first evaluate left side of the statements $$\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}:$$
$$\displaystyle\frac{{{3.56}{x}{10}^{{2}}}}{{{1.09}{x}{10}^{{4}}}}=\frac{{356}}{{10900}}\sim{0.0326}={3.26}{x}{10}^{{-{{2}}}}$$
Let us next evaluate the right side of the statements $$\displaystyle{\left({4.08}{x}{10}^{{2}}\right)}{\left({1.95}{x}{10}^{{-{{6}}}}\right)}:$$
$$\displaystyle{\left({4.08}{x}{10}^{{2}}\right\rbrace}{\left({1.95}{x}{10}^{{-{{6}}}}\right)}={\left({408}\right)}{\left({0.00000195}\right)}\sim{0.0007956}={7.956}{x}{10}^{{-{{4}}}}$$
The large the power of 10, the large the number is. This then implies that $$\displaystyle{3.56}{x}{10}^{{-{{2}}}}$$ is larger than $$\displaystyle{7.956}{x}{10}^{{-{{4}}}}$$.
$$\displaystyle\frac{{{3.56}{x}{10}^{{2}}}}{{{1.09}{x}{10}^{{4}}}}={3},{26}{x}{10}^{{-{{2}}}}{>}{7.956}{x}{10}^{{-{{4}}}}={\left({4.08}{x}{10}^{{2}}\right)}{\left({1.95}{x}{10}^{{-{{6}}}}\right)}$$
Thus the third statement $$\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}{>}{\left({4.08}{x}{10}^{{2}}\right)}{\left({1.95}{x}{10}^{{-{{6}}}}\right)}$$ is then correct.