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never [62]
3 years ago
8

I need to fine sinx=18/13

Mathematics
1 answer:
tia_tia [17]3 years ago
5 0

Answer:

Step-by-step explanation:

sin x=18/13>1

which is impossible as |sin x|≤1

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Find the following Euler Totients using Euler’s Theorem, as explained on p.409 of the text (10 points each): a.ϕ(13) b.ϕ(81) c.ϕ
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(a) \varphi(13)=12 since 13 is prime.

(b) 81=3^4, and there are 81/3 = 27 multiples of 3 between 1 and 81, which leaves 81 - 27 = 54 numbers between 1 and 81 that are coprime to 81, so \varphi(81)=54.

(c) 100=2^2\cdot5^2; there are 50 multiples of 2, and 20 multiples of 5, between 1 and 100; 10 of these are counted twice (the multiples of 2*5=10), so a total of 50 + 20 - 10 = 60 distinct numbers not coprime to 100, leaving us with \varphi(100)=100-60=40.

(d) 102=2\cdot3\cdot17; there are 51 multiples of 2, 34 multiples of 3, and 6 multiples of 17, between 1 and 102. Among these, we double-count 17 multiples of 2*3=6, 3 multiples of 2*17=34, and 2 multiples of 3*17=51; we also triple-count 1 number, 2*3*17=102. There are then 51 + 34 + 6 - (17 + 3 + 2) + 1 = 70 numbers between 1 and 102 that are not coprime to 102, and so \varphi(102)=102-70=32.

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3 years ago
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