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tatyana61 [14]
3 years ago
11

Determine the principal value of the function: Arc sin(square root of 3/2)

Mathematics
1 answer:
Drupady [299]3 years ago
5 0

Answer:

π/3

Step-by-step explanation:

We have to find the principal value of \text{arc sin}(\frac{\sqrt{3}}{2} )

arc sin means sin inverse. The sin inverse is a one to one function with its range between -\frac{\pi}{2} \textrm{ to } \frac{\pi}{2}

The principal value of the arc sin will lie within the above given range.

value of sin (60) or sin(\frac{\pi}{3}) is \frac{\sqrt{3}}{2}.

\frac{\pi}{3} lies between -\frac{\pi}{2}\textrm{ and } \frac{\pi}{2}

So, from here we can say that the Principal Value of Arc sin(square root of 3/2) is π/3

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gizmo_the_mogwai [7]

Answer: 15

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By the trapezoid midsegment theorem,

\frac{AB+CD}{2}=MN\\\frac{29+AD}{2}=22\\29+AD=44\\AD=15

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3 0
3 years ago
According to the fundamental theorem of algebra, how many zeros does the function f(x) = 4x3 − x2 − 2x + 1 have?
jarptica [38.1K]

Answer: 3 zeroes.


Explanation:-

  • The fundamental theorem of algebra states that any polynomial with degree m>0  and complex coefficients has atleast one complex root.
  • Corollary of fundamental theorem states that for any polynomial with degree m>0 has exactly m solutions.

The given function is f(x)=4x^3-x^2-2x+1

Because it is a polynomial function with degree 3>0 , Therefore by corollary of fundamental theorem of algebra , it has 3 zeroes.

8 0
3 years ago
Read 2 more answers
Soap films and bubbles are colorful because the interference conditions depend on the angle of illumination (which we aren't cov
mylen [45]

Answer:

56.39 nm

Step-by-step explanation:

In order to have constructive interference total optical path difference should be an integral number of wavelengths (crest and crest should be interfered). Therefore the constructive interference condition for soap film can be written as,

2t=(m+\frac{1}{2} ).\frac{\lambda}{n}

where λ is the wavelength of light and n is the refractive index of soap film, t is the thickness of the film, and m=0,1,2 ...

Please note that here we include an additional 1/2λ phase shift due to reflection from air-soap interface, because refractive index of latter is higher.

In order to have its longest constructive reflection at the red end (700 nm)

t_1=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_1=\frac{1}{2} .\frac{700}{(2)*(1.33)}\\ \\ t_1=131.58\ nm

Here we take m=0.

Similarly for the constructive reflection at the blue end (400 nm)

t_2=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_2=\frac{1}{2} .\frac{400}{(2)*(1.33)}\\ \\ t_2=75.19\ nm

Hence the thickness difference should be

t_1-t_2=131.58-75.19=56.39 \ nm

7 0
3 years ago
Can I PLEASE get some help?
Leviafan [203]
Well 1/40 is 0.025 so that should be the write fraction to use
4 0
3 years ago
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