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If you have a problem such as 2.56x10^4 x 4.08^2 would the numbers 2.56 and 4.08 be added or multiplied when trying to simplify it?

Then if you had an exponent that was 10^3 x 10^2 would that equal 10^5 or 10^6?

Another is 1.2 x 10^24 divided by 6.0 x 10^20 what would that equal and why?

2007-12-02 10:23:19 · 2 answers · asked by Bri 1 in Science & Mathematics Mathematics

2 answers

Scientific notation does nothing but tell you how many places the decimal place moves. For example 2.56 x 10^4

is just 2.56 x 10000 or 25,600. (see how the decimal moved four places to the right from 2.56?)

So Simplifying the first equation is 25,600 x 408 = 10,444,800 which equals 10.4448 x 10^6

10^3 = 1000 and 10^2 = 100,
1000*100=100,000=10^5, so you add the exponents together.

Third question. Works out like (1.2*6.0) x 10^(24+20)
7.2 x 10^44

2007-12-02 10:32:04 · answer #1 · answered by Dave 3 · 0 0

1. Brackets first, Unary (monadic functions with only one argument like positive and negative) next, Exponents next, Multiplication/Division next and finally Addition/Subtraction.
This is the standard order of operation.
(Computer programming languages can have their own standard order)
So: square the 4.08 first.
Then multiply the mantissas (the non-exponents).
Then scale it with the orders of magnitude (the exponents of the tens).
To multiply powers of a number you can sum the exponents of that number as long as it is exactly the same number.
e.g. 10^2 * 10^3 = 10^5
and 46^-4 * 46^11 = 46^7
not 10^4 * 11^2
2.Division is done by subtracting the exponents.
Normalizing a scientific notation number means to adjust the mantissa and exponent so that you only have one digit to the left of the decimal point.
The first thing I would do would be to de-normalize the number so that it was convenient for division i.e.
1.2e24 /6.0e20 =
12e23 / 6e20 =
2e23 /1e20 =
2e3=
2000
"e" (or "E" or "d" or "D") is just computer shorthand for scientific notation or float-point or floating point.

2007-12-02 18:54:48 · answer #2 · answered by J C 5 · 0 0

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