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Alex_Xolod [135]
3 years ago
13

Simplify sin q + cosq cotq.

Mathematics
1 answer:
IRINA_888 [86]3 years ago
3 0
\sin q+\cos q\cot q=\sin q+\cos q\cdot\dfrac{\cos q}{\sin q}=\sin q+\dfrac{\cos^2q}{\sin q}\\\\=\dfrac{sin^2q}{\sin q}+\dfrac{\cos^2q}{\sin q}=\dfrac{\sin^2q+\cos^2q}{\sin q}=\dfrac{1}{\sin q}=\csc q\\\\Used:\\\cot q=\dfrac{\cos q}{\sin q}\\\\\sin^2q+\cos^2q=1\\\\\csc q=\dfrac{1}{\sin q}
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When it is  12 o'clock the hour hand is on positive of y axis.Coordinates of the point at 12 o'clock=(0,5)

When it is 3 o 'clock the hour hand is on positive of x axis .Coordinate of the point at 3o'clock is (5,0)

When it is 6 o'clock the hour hand is on negative of y axis .The coordinates of the point at 6o'clock is (0,-5)

At 9o'clock the hour hand is on negative of x axis .The coordinate of the point at 6o'clock is(-5,0)

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3 years ago
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Tpy6a [65]

Answer:

261.5 ft^2

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The Sunday school teacher gave 5 Friend's Day invations to each of her 19 students to give to their friends. She had 5 left over
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4 years ago
Calculate the allele frequencies in this population of palm trees. A model consisting of a box containing fourteen, randomly arr
bogdanovich [222]

Answer:

The correct answer is

0.5 F, 0.5 f

Step-by-step explanation:

We note the following

Number of colored circles in the box = 14

Number of red circles in the box = 4

Number of purple circles in the box = 6

Number of blue circles in the box = 4

The allele frequency are as follows

Where the frequency is given as

Genotype            Frequency                         Relative frequency        

FF = Red                    4                                       4/14 (0.29≈0.3)                  

Ff = Purple                 6                                       6/14 (0.43≈0.4)

ff = Blue                     4                                       4/14(0.29≈0.3)

Within this population,

We however  have 4 FF = 8 F

                                6 Ff =   6 F + 6 f  and

                                4 ff = 8 f

Total allele = 8+6+6+8 = 28

Relative frequency of F = (8+6)/28 = 14/28 = 0.5

relative frequency of f = (8+6)/28 = 0.5

Therefore the allele frequencies in the palm tree population is

0.5 F, 0.5 f

When in equilibrium we have

However the FF has the product of F×F which is = F² = 0.29 so the frequency of F = √(0.29) = 0.535≈ 0.5

The frequency of Ff is Ff or fF =  0.43 since there is equal number of each allele in Ff we have fF or Ff = Ff = 0.43

Which hives 0.43/2 = F =f ≈ 0.2

To  

and ff = 0.29 so that f = 0.535 ≈ 0.5

Therefore f = F  = 0.5 + 0.2 = 0.7

5 0
3 years ago
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