How do you calculate #(5.3times10^-2) times (-2.06times10^9)#?

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Answer 1 · 2016-06-26T17:56:31

-#1.0918 xx 10^8#

This is done in the same way as you would multiply in algebra.

#3x^5 xx -4x^-2#

This could be written as#(3 xx-4) xx (x^5 xx x^-2)#
Multiply the numbers,: #3 xx-4 = -12#
Multiply the variables by adding the indices.

#x^5 xx x^-2 = x^(5-2) = x^3#

#3x^5 xx -4x^-2 = -12x^3#

We can do the same with numbers in scientific notation:
#5.3 xx10^-2 xx -2.06 xx 10^9 # can be written as:

#(5.3 xx -2.06) xx (10^-2 xx 10^9)#

Multiply the numbers,: #5.3 xx -2.06 = -10.918#

Multiply the powers by adding the indices.
#10^-2 xx 10^9 = 10^(-2+9) = 10^7#

#5.3 xx10^-2 xx -2.06 xx 10^9 = -10.918 xx 10^7#

However, in standard form there must be only one digit before the decimal point.
#-10.918 xx 10^7 = -1.0918 xx 10^8#