How do you evaluate #(2.9times10^-2)(5.2times10^-9)#?

1 Answer
Aug 2, 2017

Answer:

The 'normal' parts are multiplied and the exponents are added.
#1.5 xx10^(−10)#

Explanation:

Remember that exponents are just a way to simplify writing very large and very small numbers. They do not change the way the numbers combine in mathematics. The rules for combination do have to be observed.

#(2.9 × 10^(−2))xx(5.2 × 10^(−9))#

#(2.9 × 5.2) xx(10^(−2) × 10^(−9))#

#(15.08) xx(10^(−11))#

#1.5 xx10^(−10)#

The "longhand" version would be a simple multiplication:

#0.029 xx 0.0000000052 = 0.00000000015#