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Natasha_Volkova [10]
2 years ago
14

Please help me to get the answers please read my questions​

Mathematics
2 answers:
soldier1979 [14.2K]2 years ago
6 0

Answer: Where is the question.

Step-by-step explanation: Where is the question.

Margaret [11]2 years ago
5 0
You’re question is not on your account so where is it? If u can please comment it
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One year, a certain city experienced a
icang [17]

Answer:

Step-by-step explanation:

there is a 125 degree difference between the two

5 0
3 years ago
Help help help !!!!! iam stuck in this question
Zina [86]
It is A because x is less than one and it is open since there is no like underneath the sign.
5 0
3 years ago
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The point (2, 3) is on the terminal side of angle Θ, in standard position. What are the values of sine, cosine, and tangent of Θ
stira [4]

Answer:

sin Ф = 3/√13; cos Ф = 2/√13; and tan Ф = 3/2

Step-by-step explanation:

Let's assume we're limiting ourselves to Quadrant I.

Start with the tangent function.  tan Ф = opp / adj.

In this case opp = 3 and adj = 2.  

The length of the hypotenuse is found using the Pythagorean Theorem and is √(3² + 2²) = √13.

Then sin Ф = opp / hyp = 3/√13 or 3√13/13

and

cos Ф = adj / hyp = 2/√13 or 2√13/13

and (as before)

tan Ф =  opp / adj = 3/2

5 0
3 years ago
HELP
dsp73

Answer:

1)  4

2) 3

3) 9

4) 6

5) 3/4

Step-by-step explanation:

5 0
3 years ago
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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
ad-work [718]

(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.

4 0
4 years ago
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