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expeople1 [14]
3 years ago
9

Θ is in Quadrant III and cos^2 0= 1/4

Mathematics
1 answer:
pychu [463]3 years ago
7 0
I think you mean to say \cos^2\theta=\dfrac14. If \theta is in the third quadrant, then \sin\theta and \cos\theta. Recall that


\sin^2\theta+\cos^2\theta=1\implies\sin\theta=\pm\sqrt{1-\cos^2\theta}

We expect \sin\theta to be negative, so we take the negative root. We end up with

\sin\theta=-\sqrt{1-\dfrac14}=-\dfrac{\sqrt3}2

We also know to expect \cos\theta, so

\cos\theta=\pm\sqrt{\dfrac14}\implies\cos\theta=-\dfrac12


By definition, we have

\cot\theta=\dfrac{\cos\theta}{\sin\theta}

and so

\cot\theta=\dfrac{-\frac12}{-\frac{\sqrt3}2}=\dfrac1{\sqrt3}
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Carl is boarding a plane. He has 2 checked bags of equal weight and a backpack that weighs 4 kg. The total weight of Carl's bagg
g100num [7]

Answer:

(35 - 4)/2 = w and 15.5kg

Step-by-step explanation:

To get the weight of both bags you would have to first subtract the backpack's weight from the total weight. 35 - 4 = 31. Now divide by 2 to get the weight of one bag. 31/2 = 15.5. Each bag weighs 15.5 kg

3 0
3 years ago
Please help me I’m trying to pass this
melamori03 [73]

Answer:

I think the answer is D

Step-by-step explanation:

SOHCAHTOA

sin- Opposite and hypotenuses

Cos- Adjacent and Hypothesis

Tan- Opposite and adjacent

DE is opposite and DF is hypotenuses

7 0
3 years ago
Please help if the answer is correct you will get 5 stars :)
alukav5142 [94]

Answer:

40

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Test the series for convergence or divergence (using ratio test)​
Triss [41]

Answer:

    \lim_{n \to \infty} U_n =0

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given series is an alternating series

∑(-1)^{n} \frac{n^{2} }{n^{3}+3 }

Let   U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }

By using Leibnitz's rule

   U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }

 U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}

Uₙ-Uₙ₋₁ <0

<u><em>Step(ii):-</em></u>

    \lim_{n \to \infty} U_n =  \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }

                       =  \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }

                    = =  \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }

                       =\frac{1}{infinite }

                     =0

    \lim_{n \to \infty} U_n =0

∴ Given series is converges

                       

                     

 

3 0
3 years ago
I can't figure this one out...
worty [1.4K]

Answer:

the area would be 36(pi) since you can't find out the exact value of pi (22/7) and where it would terminate so thats the most exact answer you can get. The circumference would be

Step-by-step explanation:

12/2 = 6

Area = (pi)radius^2

6^2 = 36

36(pi)

Circumference: (pi)d

12(pi)


For approximate area you multiply r^2 by 3.14

For approximate circumference you multiply diameter by 3.14


5 0
3 years ago
Read 2 more answers
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