SOLUTION: Tutor I am trying to learn how to do scientific notations: Could you help me please? (4.2*10^-2) (3.5*10^8) would this be 4,200,000,000 times 3,500,000, how would I compute th

Algebra ->  Decimal-numbers -> SOLUTION: Tutor I am trying to learn how to do scientific notations: Could you help me please? (4.2*10^-2) (3.5*10^8) would this be 4,200,000,000 times 3,500,000, how would I compute th      Log On


   



Question 135027: Tutor I am trying to learn how to do scientific notations: Could you help me please?
(4.2*10^-2) (3.5*10^8) would this be 4,200,000,000 times 3,500,000, how would I compute this and write it in scientific notation?

also 4.2*10^-6/7*10^8 4.2.10^-6/7.10^8 I don't know how to do this one or where do I start?
Karen
thank you for your help!!!

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Given : (4.2*10^-2) (3.5*10^8)
10%5E-2 = .0.01
Think of it this way. 10%5E0 = 1
10^1 = 10
10^2 = 100
10^-1 = 0.1
10^-2 = 0.01
You can see that each number has the decimal point moved. For positive numbers, the decimal point moves one place to the right with each increase of 1. With negative numbers, the decimal moves one place to left with each decrease of 1.
For scientific notation, the "number part" is always between 1 and 10.
So
%284.2%2A10%5E-2%29+ is already in scientific notation. So is %283.5%2A10%5E8%29+

Use the same technique to solve the second one. Ping me if you can't get it
The value of %284.2%2A10%5E-2%29+ is 0.042 (decimal moved two place to the left.
The value for %283.5%2A10%5E8%29 is 350,000,000 (deicmal 8 place to the right)
To multiply the numbers, use the scientific notation form.
%284.2%2A10%5E-2%29%283.5%2A10%5E8%29
%284.2%29%2A%283.5%29+%2A+10%5E%28-2%2B8%29
14.7+%2A+10%5E6
You are almost there now. The only thing left to do is make the 'number part' be between 1 and 10.
1.47%2A10%5E1+%2A+10%5E6
1.47%2A10%5E7

Use the same technique for the second one. Just remember that when you divide, that is the same as subtracting exponents. Ping me by email if you can't get it.