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N76 [4]
3 years ago
6

Sabendo que senx=3/5 e que X € 1°quadrante, calcule tgx

Mathematics
1 answer:
guapka [62]3 years ago
4 0
<span>se 
sen x = 3/5 eleva ao quadrado ambos os lados 
sen² x = (3/5)² 

Tem que partir da relação fundamental 

sen²x + cos²x = 1 
(3/5)² + cos² x = 1 

cos² x = 1 - (3/5)² = 16/25 

cos x = V(16/25 = 4/5 

b ) tg x = sen x / cos x = 3/5 / 4/5 = 3/5 * 5/4 = 3/4 

c) cotg x = cosx / sen x = 1 / tgx = 1/ (3/4 = 4/3 

d) sec x = 1/ cos x = 1 / 4/5 = 5/4 

e) cossec x = 1 / sen x = 1/ 3/5 = 5/3 

se gostou , avalie 
edson</span>edson <span>· 9 anos atrás</span>
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Answer:

3

Step-by-step explanation:

Triangle ABO is a right triangle, and so we can use the Pythagorean Theorem to find the radius, r:

r² + 6² = 7.5²,  or  r² = 56.25 - 36 = 20.25

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4 years ago
You have quarters and nickels saved in a piggy bank. There is a total of $3.45 in the bank. If you have 70 cents in nickels, how
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11 quarters

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Two chess players, A and B, are going to play 7 games. Each game has three possibleoutcomes: a win for A (which is a loss for B)
Lina20 [59]

Answer:

A) the possible outcomes for individual game is 210 games.

B) The possible outcome for this is 357 games.

C) Then the Possible Games are 267 games.

Step-by-step explanation:

A) Total number of individual games are 7 in which A ends up with 3 wins which give 4 remaining games. then there are two draw games from them. and in the end the remaining games are losses so,we useformula for combination:

                       (7C3)*(4C2)=210 \ games

B) Now Player A has 4 point that gives us 5 possibilities until we reach 7 games. Similarly B have 3 points which is same as player A gives us 5 possibilities until we reach 7 games. thus those two cases for player A and Player B can be stated as:

           (7C3)+(7C3)*(4C2)+(7C1)*(6C6)+(7C2)*(5C4)=357\  games

C) Lets say that player A ahs 4 points and player B has 3 points and all seven games have been played so.

c1) If player A has 4 wins and 3 losses then last win have to be in 7th match thus:the answer is (6C3).

c2) If player A has 3 win, 2 draws and 3 losses thus it means that final match cannot be a loss thus the answer is (6C2)*(5C3).

c3) Now lastly if player A has 1 win and 5 draws we can arrange them arbitrarily thus the answer here is (7C1).

c4) now If player A has 2 wins, 4 draws and 2 losses thus answer is (6C1)*(6C2)

Sum of all the cases is

               (6C3)+(6C2)*(5C3)+(7C1)+(6C1)*(6C2)=267\ games

4 0
3 years ago
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