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hram777 [196]
3 years ago
11

Please help i have 1 question. thank you.

Mathematics
1 answer:
liberstina [14]3 years ago
6 0
Since the direction is not specified, it can be safe to assume that the direction of rotation is counterclockwise.

300 degrees is 60 degrees less than 360 degrees, which is a full rotation. Thus, 300 degrees counterclockwise is the same as 60 degrees clockwise.

Note that this shape is a hexagon. Thus, we can divide a hexagon into 6 equilateral triangles, each with measures of 60 degrees. Just move OQ to the adjacent clockwise equilateral triangle and see what it overlaps with.

OQ is the altitude of the equilateral triangle, so our answer will be the altitude of the adjacent clockwise equilateral triangle.

The answer is OF.

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Answer:

Step-by-step explanation:

Her angle of depression is the same measure as the angle of inclination inside the triangle. That's the base angle that is not the right angle. The height of the triangle is the 10,000 feet she is up in the air, and the reference angle is 20, and we are looking for the measure of the base of this right triangle. The trig ratio that uses the side opposite over the side adjacent is the tangent ratio:

tan(20)=\frac{10000}{x} and

x=\frac{10000}{tan20} so

x = 27474.77 feet

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3 years ago
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Step-by-step explanation:

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3 years ago
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What is the solution to 2x^2+8=x^2-16?
KIM [24]
X= 2i(square root of 6), -2i(square root of 6)
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4 years ago
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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
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