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Hitman42 [59]
3 years ago
11

What is the remainder when the product of the 5 smallest prime numbers is divided by 42?

Mathematics
1 answer:
sveta [45]3 years ago
5 0

Answer:

0

Step-by-step explanation:

The five smallest prime numbers are 2, 3, 5, 7 and 11.

2 × 3 × 5 × 7 × 11

= 2310

Divide by 42.

2310/42

= 55, Remainder 0

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If r=20.5 and s=34.2 find S Round to the nearest tenth
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Step-by-step explanation:

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Then, you get that the measure of "S" rounded to the nearest tenth is:

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7 0
4 years ago
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8 0
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