Quicker Way To Do Standard Form?

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Y2_UniMaths
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#1
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#1
Is there another method to answer these questions instead of multiplying everything out and going the long way? Like you can with addition and subtraction?
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gdunne42
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#2
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(Original post by Y11_Maths)
Is there another method to answer these questions instead of multiplying everything out and going the long way? Like you can with addition and subtraction?
E.g,
(1.2 x 10^8) x 2000 =
(1.2 x 10^8) x (2 x 10^3) =
(1.2 x 2) x (10^8 X 10^3) then do the x and use rules of indices on the powers of 10

(6 x 10^5) divided by (2 x 10^3) =
(6/2) x (10^5/10^3) then do the divide and use rules of indices
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dsmith23
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#3
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#3
If you multiply 7.2 by 3, what do you get? If you do that calculation can you see how you would be able to get to the final answer?
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Y2_UniMaths
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#4
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(Original post by dsmith23)
If you multiply 7.2 by 3, what do you get? If you do that calculation can you see how you would be able to get to the final answer?
Yes that is a very easy way thanks
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Y2_UniMaths
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#5
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#5
(Original post by gdunne42)
E.g,
(1.2 x 10^8) x 2000 =
(1.2 x 10^8) x (2 x 10^3) =
(1.2 x 2) x (10^8 X 10^3) then do the x and use rules of indices on the powers of 10

(6 x 10^5) divided by (2 x 10^3) =
(6/2) x (10^5/10^3) then do the divide and use rules of indices
On the second one why does the divide change to multiply after you swap the numbers?
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gdunne42
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#6
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(Original post by Y11_Maths)
On the second one why does the divide change to multiply after you swap the numbers?
To divide two things in standard form you divide the numbers by each other and divide the powers of 10 by each other to produce the standard form result. The times was always there, the "swap" you refer to is just rearranging the original calculation using the rules of maths.
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Y2_UniMaths
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#7
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#7
(Original post by gdunne42)
To divide two things in standard form you divide the numbers by each other and divide the powers of 10 by each other to produce the standard form result. The times was always there, the "swap" you refer to is just rearranging the original calculation using the rules of maths.
Ok cheers
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