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Natali [406]
3 years ago
14

How many six digit even numbers are possible if the leftmost digit cannot be zero

Mathematics
1 answer:
Schach [20]3 years ago
4 0

1st digit cannot be 0 so there are 9 choices {1,2,3,4,5,6,7,8,9}. The middle 4 digits have 10 choices (can have 0). The last digit must be EVEN or 0 to make the entire 6 digit number even, {0,2,4,6,8} = 5 choices

9×10×10×10×10×5 = 450,000
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finlep [7]

For this case we have the following expression:

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(\frac {-27y ^ {- 2}} {54x ^ {- 5} * y ^ {- 4}}) ^ {- 2} =

Simplifying:

(\frac {-y ^ {- 2}} {2x ^ {- 5} * y ^ {- 4}}) ^ {- 2} =

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So, rewriting the expression we have:

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SImplifying:

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\frac{4}{x^{10}y^{4}}

8 0
4 years ago
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