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Alja [10]
3 years ago
10

How many five-digit codes can be made if the first number cannot be a 0 or a 1 and no numbers can repeat?

Mathematics
1 answer:
Makovka662 [10]3 years ago
6 0

Answer:

24,192.

Step-by-step explanation:

The first number must be one of 2,3,4,5,6,7,8 or 9. That is 8 possibilities.

The number of permutations of the other 4 numbers is 9P4

= 9! /  (9-4)!

= 3024

Now we multiply by the 8:

3024 * 8

= 24,192.

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dolphi86 [110]
Answer: Choice C
7 divided by cosec 50 degrees

=====================================

Explanation:

Abbreviations:
sec = secant
cot = cotangent
cosec = cosecant (csc is also a widely used abbreviation)

-----------------

Let the horizontal leg be x

This leg is opposite the angle 50 degrees. The hypotenuse is 7

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We'll either use the sine rule or the cosecant (csc) rule
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Since "sine" is nowhere to be found in the answer choices, we'll use the cosecant rule

-----------

cosecant = hypotenuse/opposite
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Answer:

Step-by-step explanation:

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