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3 years ago

12

Write the equation using polar coordinates y^2=16x

1 answer:

baherus [9]3 years ago

8 0

<span>r²sin²θ = 16rcosθ </span>

<span>rsin²θ = 16cosθ </span>

<span>r = 16cosθ / sin²θ </span>

<span>r = 16cotθcscθ</span>

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the point (3,-6) lies on the terminal side of angle 0. Find the exact value of the six trigonometric functions of 0. Sin cos tan

Shkiper50 [21]

ANSWER

See explanation

EXPLANATION

The point (3,-6) is the fourth quadrant.

In this quadrant only the cosine ratio and the secant ratio are positive.

The remaining four trigonometric ratios are negative.

The diagram is shown in the attachment.

We use the Pythagoras Theorem to find the hypotenuse, h.

${h}^{2} = {6}^{2} + {3}^{2}$

${h}^{2} = 36 + 9$

${h}^{2} = 45$

$h = \sqrt{45}$

$h = 3 \sqrt{5}$

$\sin( \theta) = - \frac{opp}{hyp}$

$\sin( \theta) = - \frac{6}{3 \sqrt{5} }$

$\sin( \theta) = - \frac{2 \sqrt{5} }{ 5 }$

$\sin( \theta) = - \frac{2}{ \sqrt{5} }$

$\ cos(\theta) = \frac{adj}{hyp}$

$\ cos(\theta) = \frac{3}{3 \sqrt{5} }$

$\ cos(\theta) = \frac{1}{ \sqrt{5} }$

$\ cos(\theta) = \frac{ \sqrt{5} }{5}$

$\tan(\theta)= - \frac{opp}{hyp}$

$\tan(\theta)= - \frac{6}{3} = - 2$

$\csc(\theta)= - \frac{hyp}{opp}$

$\csc(\theta)= - \frac{3 \sqrt{5} }{6} = - \frac{\sqrt{5} }{2}$

$\sec( \theta) = \frac{hyp}{adj}$

$\sec( \theta) = \frac{3 \sqrt{5} }{3} = \sqrt{5}$

$\cot( \theta) = - \frac{adj}{opp}$

$\cot( \theta) = - \frac{3}{6} = - \frac{1}{2}$

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3 years ago

What is the equation of the function after is reflected over the x-axis f(x)=(x+4)^3

Triss [41]

Answer:

f(x) = -(x+4)^3

Step-by-step explanation:

Reflection over the x-axis negates every y-coordinate. The reflected function is ...

f(x) = -(x+4)^3

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