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forsale [732]
3 years ago
7

Find all solutions in the region [0,2pi) for the equation cos^2(2x)-sin^2(2x)=0

Mathematics
1 answer:
Lynna [10]3 years ago
4 0
\cos^{2}(2x)-\sin^{2}(2x)=0
\rightarrow \sin^{2}(2x)=\cos^{2}(2x)
\rightarrow \frac{\sin^{2}(2x)}{\cos^{2}(2x)}=1
\rightarrow \tan^{2}(2x)=1
\rightarrow \tan(2x)=\pm 1

2x can range anywhere in [0, 4\pi)
So:
\rightarrow 2x=\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{9\pi}{4}, \frac{11\pi}{4}, \frac{13\pi}{4}, \frac{15\pi}{4}
\rightarrow x=\frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8}, \frac{15\pi}{8}
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A, B and C are partners sharing profits in the ratio 3:2:1. During the year A withdraws Rs. 3000 at the start of each month, B w
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Answer:

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Step-by-step explanation:

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5 0
3 years ago
The Los Angeles Times regularly reports the air quality index for various areas of Southern California. A sample of air quality
egoroff_w [7]

Answer:

1. (a) 32     (b) 8

2. (a) 48.33    (b) 92.75    (c) 9.63

3.  We observe that sample mean of air quality index in Pomona is higher that that of Anaheim which indicates that the level of air pollution is higher in Pomona and there is more health concern here.

On the other hand, the variation in air quality index is higher in Anaheim than in Pomona as Anaheim has higher variance and standard deviation for air quality index which means that there are not much fluctuations in air quality index in Pomona.  

Step-by-step explanation:

First arranging our data of air quality index values for Pomona in ascending order, we get : 28, 42, 45, 48, 49, 50, 55, 58, 60.

  1. (a) Range is given by the formula :

                      Highest value in data - Lowest value in data = 60 - 28 = 32

        (b) Interquartile rage = Third quartile - First quartile

                                           = Q_3 - Q_1

             Q_1 =  (\frac{n-1}{4} )^{th} observation in the data = 2^{nd} obs. { Because n = 9}

                 Therefore, Q_1 = 42.

             Q_3 = (3Q_1)^{th} obs. = (3 x 2 ) = 6^{th} obs = 50

             So, Interquartile rage = 50 - 42 = 8.

     2. (a) Sample mean formula =  \frac{\sum X_i}{n} = \frac{28+ 42+ 45+ 48+ 49+ 50+ 55+ 58+ 60}{9} = 48.33

         (b) Sample variance formula = \frac{\sum (X_i - Xbar)^{2}}{n-1} where Xbar = Sample mean

                                                         = 92.75

          (c) Sample standard deviation  formula = \sqrt{variance} = \sqrt{92.75} = 9.63.

     3. Given that a sample of air quality index readings for Anaheim has   sample mean of 48.3, a sample variance of 136, and a sample standard deviation of 11.66.

We observe that sample mean of air quality index in Pomona is higher that that of Anaheim which indicates that the level of air pollution is higher in Pomona and there is more health concern here.

On the other hand, the variation in air quality index is higher in Anaheim than in Pomona as Anaheim has higher variance and standard deviation for air quality index which means that there are not much fluctuations in air quality index in Pomona.  

                                       

7 0
3 years ago
I’ll give brainly <br> Solve for x.<br> 27<br> 36<br> x<br> 28<br> X= <br> ?
alexandr402 [8]

Answer:

i dont understan the method you gave me can youbplease ask again showing a different method

6 0
3 years ago
Bella saw that 7 out of 11 people brought an umbrella to the beach. There 99 people there. How many of them did not have a beach
Alinara [238K]
Hello!

7 out of 11 people brought an umbrella to the beach

this is a fraction of 7/11

We multiply this by 99 to get how many people brought an umbrella out of 99

7/11 * 99 = 63

You subtract this from 99 to see how many did not have an umbrella

99 - 63 = 36

The answer is 36 people

Hope this helps!
7 0
3 years ago
Is a three-digt whole number
Rudiy27

Answer:

693

Step-by-step explanation:

7*9*11*3=693

5 0
3 years ago
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