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stepladder [879]
3 years ago
8

An=-31+(n-1)-11 find the eleventh term

Mathematics
1 answer:
Lilit [14]3 years ago
3 0

An=-31+(n-1)-11 find the eleventh term

a11 = -31 + (-11) (11-1)

a11 = -31 +-11(10)

a11 =-31 +-110

a11 = -142

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Firdavs [7]

Answer:

sin\theta_1 =  - \frac{2\sqrt{70}}{19}

Step-by-step explanation:

We are given that \theta_1 is in <em>fourth</em> quadrant.

cos\theta_1 is always positive in 4th quadrant and  

sin\theta_1 is always negative in 4th quadrant.

Also, we know the following identity about sin\theta and cos\theta:

sin^2\theta + cos^2\theta = 1

Using \theta_1 in place of \theta:

sin^2\theta_1 + cos^2\theta_1 = 1

We are given that cos\theta_1 = \frac{9}{19}

\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 =  \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 =  \dfrac{280}{361}\\\Rightarrow sin\theta_1 =  \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 =  +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}

\theta_1 is in <em>4th quadrant </em>so sin\theta_1 is negative.

So, value of sin\theta_1 =  - \frac{2\sqrt{70}}{19}

6 0
2 years ago
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Dafna11 [192]
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3 years ago
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Answer:

B

Step-by-step explanation:

The options in the question and the number line are shown in the image attached.

From the options, it is clear that the difference between the temperatures at 7:00 a.M. and 1:00 p.M is;

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