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

Answer 1

To evaluate (5.18 × 10^2) × (9.1 × 10^-5), you first multiply the numbers outside the parentheses: 5.18 × 9.1. Then, you add the exponents of 10: 2 + (-5) = -3. So, the result is (5.18 × 9.1) × 10^-3. Finally, calculate the product of 5.18 and 9.1, which is approximately 47.138, yielding the final result of 4.7138 × 10^-2.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Genesis Allen

#0.047138#

#(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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 3

Hazel Anglin

#4.7138xx10^(-2)#

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)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.