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3 years ago

Find the exact value of cot(11pi/6) cos(11pi/6) csc(11pi/6)

1 answer:

5 0

A) cot ( 11π/6 ) = cot ( 2π - π/6 ) = cot ( - π/6 ) = -√3
B ) cos ( 11π/6 ) = cos ( -π/3 ) = √3/2
C ) csc ( 11π/6 ) = 1 / sin ( 11π/6 ) = 1/ sin ( -π/6 ) = 1/(-1/2) = - 2

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pashok25 [27]

Answer:

Step-by-step explanation:

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2x²=800

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3 years ago

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makkiz [27]

Answer:

14 is the length of the missing side.

Step-by-step explanation:

Use pythagorean theorem.

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3 years ago

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Step-by-step explanation:

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3 years ago

-7x - 2y = 14 6x + 6y = 18 x = y =

Mrac [35]

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Step-by-step explanation:

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3 years ago

Use l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Ove

steposvetlana [31]

Answer:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}$

Step-by-step explanation:

The limit is:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}$

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

$\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}$

by replacing, and applying repeatedly you obtain:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}$

hence, the limit of the function is -1/14

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3 years ago