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This is an exam question, for Grade 11 high scool students.

 

How would you go about solving this ? cheeky

Guest Nov 3, 2015

Best Answer 

 #1
avatar+92191 
+15

\(2^{2015}\;*\;5^{2019}\\ =2^{2015}\;*\;5^{2015+4}\\ =2^{2015}\;*\;5^{2015}\;*\;5^{4}\\ =(2*5)^{2015}\;*\;5^{4}\\ =10^{2015}\;*\;625\\ =625\;*\;10^{2015}\\ =6.25*10^2\;*\;10^{2015}\\ =6.25*\;10^{2017}\\\)

 

That is how I would solve it    wink   cheeky

Melody  Nov 3, 2015
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1+0 Answers

 #1
avatar+92191 
+15
Best Answer

\(2^{2015}\;*\;5^{2019}\\ =2^{2015}\;*\;5^{2015+4}\\ =2^{2015}\;*\;5^{2015}\;*\;5^{4}\\ =(2*5)^{2015}\;*\;5^{4}\\ =10^{2015}\;*\;625\\ =625\;*\;10^{2015}\\ =6.25*10^2\;*\;10^{2015}\\ =6.25*\;10^{2017}\\\)

 

That is how I would solve it    wink   cheeky

Melody  Nov 3, 2015

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