How do you evaluate #(2.05times10^-8)/(4times10^-2)#?

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Answer 1 · 2017-04-20T16:03:12

#= 5.125 xx 10^-7#

Do not be tempted to change the numbers into decimal notation - that defeats the purpose of using scientific notation in the first place.

Consider #" "(12x^-5)/(4x^-3)#

This could be written as: #(12 x ^(-5-(-3)))/4#

Simplifying leads to:#" "3 xx x^-2 = 3/x^2#

In the same way #" "(2.05 xx 10^-8)/(4 xx 10^-2#

can be written as #" "(2.05 xx10^(-8-(-2)))/4#

#=0.5125 xx 10^-6#

Write in correct scientific notation - move the decimal place one place to the right the index will decrease.

#= 5.125 xx 10^-7#