How do you simplify and write #0.0007 xx 190# in scientific notation?
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See a solution process below:
First, write each term in scientific notation:
For #0.0007# we need to move the decimal point 4 places to the right therefore the exponent of the 10s term will be negative:
#0.0007 = 7.0 xx 10^-4#
For #190# we need to move the decimal point 2 places to the left therefore the exponent of the 10s term will be positive:
#190 = 1.9 xx 10^2#
We can now rewrite this expression as:
#0.0007 xx 190 => (7.0 xx 10^-4)(1.9 xx 10^2) => #
#(7.0 xx 1.9)(10^-4 xx 10^2) => 13.3(10^-4 xx 10^2)#
We can now use this rule of exponents to combine the 10s terms:
#x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) + color(blue)(b))#
#13.3(10^color(red)(-4) xx 10^color(blue)(2)) => 13.3 xx 10^(color(red)(-4) + color(blue)(2)) => 13.3 xx 10^-2#
To put this in true scientific notation we need to move the decimal point one place to the left so we need to add #1# to the 10s exponent:
#13.3 xx 10^-2 => 1.33 xx 10^(-2 + 1) => 1.33 xx 10^-1#
A lot of detail provided to help with understanding.
#"1.33xx10^(-1)#
Note that #10^0=1" and that "10^1=10#
#0.0007->0.0007xx1/10^0#
#0.0007->0.007xx1/10^1#
#0.0007->0.07xx1/10^2#
#0.0007->0.7xx1/10^3#
#0.0007->7.0xx1/10^4#
..............................................................................
#190->190xx10^0#
#190->19.0xx10^1#
...........................................................................
So we can write: #" "0.0007xx190# as
#7xx19xx1/10^4xx10#
Not that #7xx19# is the same as #(7xx20)-7=140-7=133# giving:
#133xx1/10^4xx10" "->133xx1/10^3#
but #133" is the same as "1.33xx10^2# giving:
#1.33xx10^2/10^3#
#1.33xx1/10^" "->" "1.33xx10^(-1)#
