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OLEGan [10]
3 years ago
14

How many area codes (ABC) would be possible if the first 2 digits can be any number 2-9 & last digit can be any number 0-9?

Mathematics
1 answer:
Korolek [52]3 years ago
8 0
The answer would be A. 640

Possibilities for the first two are 8 and 8 (2-9 is 8 numbers) and the last is 10 (0-9 is 10 numbers)

So the probability is found by multiplying 8*8*10 which is 640
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OLEGan [10]

Answer:

Step-by-step explanation:

1). x² - 8x

  To convert this expression into a perfect square trinomial,

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  Option (4) is the answer.

2). x² + 2x = 3

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     x - 3 = ±4

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