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igor_vitrenko [27]
4 years ago
9

What is an equation for a sine curve with amplitude 2, and period 4pi radians ?

Mathematics
1 answer:
Nezavi [6.7K]4 years ago
8 0
<span>1. a sine curve with amplitude 2, and period 4pi radians
</span>
the general equation of the sine curve ⇒⇒ y = a sin (nθ)
where: a is the amplitude  and  n = 2π/perid
∵ <span>amplitude 2, and period 4pi radians
</span>
∴ y = 2 sin (θ/2)

The correct answer is option D. y = 2 sin (θ/2)
===========================================

<span>2.The period and amplitude of the function ⇒⇒ y = 5 cos 2θ
</span>
<span>comparing with y = a cos nθ
</span>
where : a is the amplitude  and  n = 2π/period
<span>amplitude = 5  , period = 2π/n = 2π/2 = π
</span>

The correct answer is option B. Period: pi radians: Amplitude:5

============================================================
3. tan (2π/3) = tan 120° = -√3 
120° lie in the second quadrant and its reference angle = 180° - 120° = 60°
tan function in the second quadrant is negative
∴ tan 120° = - tan 60 = -√3

The correct answer is C. -sqrt3

=====================================================
4. <span>Tan 5π/6 = tan 150° = -(√3)/3
</span>
150° lies in the second quadrant and its reference angle = 180° - 150° = 30°
tan function in the second quadrant is negative
∴ tan 150° = - tan 30 = -(√3)/3

The correct answer is <span>B.-sqrt3/3</span>
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Step-by-step explanation:

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__

Suppose we choose a=2 and b=-1 to create a square root curve that starts to the right of the y-axis and opens to the right. (The "left shift" is negative.) For these values, we will get an integer solution when 2x-1 is an odd square. That is, suitable x-values could be 1, 5, 13, 25, and corresponding y-values would be 1, 3, 5, 7.

Any line that passes through any of these points, but does not cross the square root curve again, will do. The one I proposed in the attachment goes through the point (5, 3) and has a slope of 2. (c=2, d=-7)

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3 years ago
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