How do you evaluate #(5.18times10^2)(9.1times10^-5)#?

2 Answers
Jul 18, 2017

Answer:

#0.047138#

Explanation:

#(5.18 xx 10^2)(9.1 xx 10^-5)#

#:.a^-5=1/a^5#

#:.=(5.18 xx 100)(9.1/10^5)#

#:.=518 xx 9.1/100000#

#:.=518 xx 0.00091#

#:.=0.047138#

Jul 18, 2017

Answer:

#4.7138xx10^(-2)#

Explanation:

Lets get rid of all the decimal places and put them back at the end.

#5.18xx10^2=518#

#9.1xx10^(-5)=91xx10^(-6)#

So now we have: #518xx91xx10^(-6) larr" "10^(-6)" is the same as "1/10^6#

For the moment disregard the #10^(-6)#

#color(white)(00)518#
#ul(color(white)(000)91)larr" Multiply"#
#color(white)(0)4662larr 518xx90#
#ul(color(white)(00)518)larr 518xx1#
#47138 larr 4662+518#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now we put back the #xx1/10^6#

#47138xx1/10^6" "=" "0.047138#

Writing this in the same format as in the question (a good move)

#4.7138xx1/10^2" "->" " 4.7138xx10^(-2)#