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alekssr [168]
3 years ago
10

Determine the quadrant where θ lies given that sinθ < 0 and tanθ > 0.

Mathematics
2 answers:
tino4ka555 [31]3 years ago
4 0
The answer to this question is quadrant III. Since
sinθ = y/r and tanθ = y/x, we have, in quadrant III, x, y negative.

sinθ < 0 (r always positive) but tanθ > 0 (negative divided by negative is positive).
cricket20 [7]3 years ago
3 0

Answer:

Option c. is correct.

Step-by-step explanation:

if sin\theta < 0 means values of sin\theta will be negative.

For tan\theta =\frac{sin\theta }{cos\theta } > 0 means  positive value of tan\theta

values of sin\theta and cos\theta  both should be either > 0 or < 0 for the poitive values of tan\theta. If any one, sine or cosine is negative then tan\theta will be < 0

We know that for the condition tan\theta > 0 and sin\theta < 0 value of cos\theta should be negative.

Therefore, \theta should lie in 3rd quadrant where sin\theta and cos\theta both are negative.

Option C. 3rd option is correct.

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Maya, Eric, and Ryan sent a total of 116 text messages during the weekend. Ryan sent 3 times as many messages as Maya. Eric sent
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Step-by-step explanation:

m+e+r=116

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Which of the following is equivalent to 5 cube root 13^3
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Suppose that we have a sample space S5 {E1, E2, E3, E4E4E, E5, E6, E7}, where E1, E2, . . ., E7 denote the sample points. The fo
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Answer:

P(A) = 0.4 ; P(B) = 0.50 ; P(C) = 0.60 ; P(A u B) = 0.65 ; P(A n B) = 0.25 ;

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0.5

Step-by-step explanation:

S = {E1, E2, E3, E4, E5, E6, E7}

P(E1) = .05, P(E2) = .20, P(E3) = .20, P(E4) = .25, P(E5) = .15, P(E6) = .10, and P(E7) = .05

Let:

A{E1,E4,E6}

B{E2,E4,E7}

C{E2,E3,E5,E7}

P(A) = 0.05 + 0.25 + 0.10 = 0.40

P(B) = 0.20 + 0.25 + 0.05 = 0.50

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P(A u B) = 0.05 + 0.20 + 0.25 + 0.10 + 0.05 = 0.65

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A n C = ∅

Hence, A and C are mutually exclusive.

B complement = B' = {E1, E3, E5, E6}

P(B') = 0.05 + 0.20 + 0.15 + 0.10 = 0.5

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