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
ANSWERED
asked 2020-10-31
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>
\(\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}{>}{\left({4.08}\cdot{10}^{{2}}\right)}{\left({1.95}\cdot{10}^{{-{{6}}}}\right)}\)
\(\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}={\left({4.08}\cdot{10}^{{2}}\right)}{\left({1.95}\cdot{10}^{{-{{6}}}}\right)}\)

Answers (1)

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.
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