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1 answer:
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Answer:
Circular paraboloid
Step-by-step explanation:
Given ,
Here, these are the respective axes components.
- <em>Component along x axis
</em>
- <em>Component along y axis
</em>
- <em>Component along z axis
</em>
We see that , from the parameterised equation ,
This can also be written as :
This is similar to an equation of a parabola in 1 Dimension.
By fixing the value of z=0,
<u><em>We get which is equation of a parabola curving towards the positive infinity of y-axis and in the x-y plane.</em></u>
By fixing the value of x=0,
<u><em>We get which is equation of a parabola curving towards positive infinity of y-axis and in the y-z plane. </em></u>
Thus by fixing the values of x and z alternatively , we get a <u>CIRCULAR PARABOLOID. </u>
Answer:
- 150° and 210°
- 0
Step-by-step explanation:
It is convenient to memorize the trig functions of the "special angles" of 30°, 45°, 60°, as well as the way the signs of trig functions change in the different quadrants. Realizing that the (x, y) coordinates on the unit circle correspond to (cos(θ), sin(θ)) can make it somewhat easier.
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<h3>20.</h3>You have memorized that cos(x) = (√3)/2 is true for x = 30°. That is the reference angle for the 2nd-quadrant angle 180° -30° = 150°, and for the 3rd-quadrant angle 180° +30° = 210°.
Cos(x) is negative in the 2nd and 3rd quadrants, so the angles you're looking for are
150° and 210°
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<h3>Bonus</h3>You have memorized that sin(π/4) = √2/2, and that cos(3π/4) = -√2/2. The sum of these values is ...
√2/2 + (-√2/2) = 0
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<em>Additional comments</em>
Your calculator can help you with both of these problems.
The coordinates given on the attached unit circle chart are (cos(θ), sin(θ)).
