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Mars2501 [29]
3 years ago
13

If the measure of one exterior angle of a regular polygon is 156 degrees , how many sides does the polygon have ?

Mathematics
1 answer:
Burka [1]3 years ago
8 0

Answer:

15  sides

Step-by-step explanation:

with regular polygons it is best to work with the exterior angles since the sum of exterior angles is  360  degrees

one exterior angle =  180  − one interior angle

∴  one ext. <  =  180  −  156 =  24  degrees

no of sides  = 360 /24 =  15  s i d e s

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A student randomly draws a card from a standard deck and checks to see if it is his favorite suit he then returned to card to th
tensa zangetsu [6.8K]

Answer:

The binomial probability formula can not be used for this experiment because it does not state the number of times he expects to draw his favorite suit.  

Step-by-step explanation:

The binomial probability formula is expressed as follows:

P (k success in n trials) = \left \{ {{n \atop k} \right. p^{k}q^{n-k}

n = number of trials, k = number of successes, n-k = number of failures, p = probability of success in one trial and q = 1 - p = probability of failure in one trial.  

In the given problem, all of the variables are known except for 'k', the amount of times that the student predicts he will draw his favorite suit.  

8 0
4 years ago
I'll give brainliest!!!!!!
ludmilkaskok [199]
-2,2 is the answer hoped it helped :)
7 0
4 years ago
Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20
Mice21 [21]

Power and chain rule (where the power rule kicks in because \sqrt x=x^{1/2}):

\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'

Simplify the leading term as

\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}

Quotient rule:

\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}

Chain rule:

(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)

(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)

Put everything together and simplify:

\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}

=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}

=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}

5 0
3 years ago
I don't understand this problem...like how to combine the k terms or how to get the answer. ​
creativ13 [48]

Answer:

-6 k +5

Step-by-step explanation:

-2k -4k +5

Think of k is apples

-2 apples -4 apples

-6 apples

-6 k +5

3 0
3 years ago
Read 2 more answers
Find the missing factor 7s^2+25s+12=(s+3)()
Deffense [45]
7s^2+25s+12=\\&#10;7s^2+21s+4s+12=\\&#10;7s(s+3)+4(s+3)=\\&#10;(s+3)(7s+4)

It's 7s+4.
8 0
3 years ago
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