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Hitman42 [59]
4 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]4 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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Given: sin theta=2/3 and is in the second quadrant; evaluate the following expression. sin 2 theta
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Step-by-step explanation:

Process

1.- Find the cos a

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  Hypotenuse = 13

  Adjacent side = ?

  Adjacent side² = 13² - 5²

                          = 169 - 25

                          = 144

 Adjacent side = 12

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2.- Find the sin b

sin b = 5/13

3.- Use the formula below to find sin(a + b)

              sin (a + b) = sina cosb  + sin b cos a

- Substitution

              sin (a + b) = (5/13)((-12/13) + (5/13)/12/13)

- Simplification

              sin (a + b) = -60/169 + 60/13

- Result

              sin(a + b) = 0                        

Step-by-step explanation:

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3 years ago
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3 years ago
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