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iVinArrow [24]
3 years ago
8

The graph of the sine curve below is of electromagnetic energy that represents green light:

Mathematics
1 answer:
Gre4nikov [31]3 years ago
4 0

Answer:

y=sin((pi/290)x)

Step-by-step explanation:

If we look at all the coordinates given (0,0),(145,1),(290,0),(435,-1),(580,0) we can draw it out and see what are our maximum and minimum y-value.  Once you graph this you will notice that the maximum y-value is 1 and minimum y-value is -1, this is important because we call this our amplitude.  Now we will determine our midline, we do so by getting the average of our amplitude so \frac{1+(-1)}{2}=\frac{0}{2} =0 and so our midline is zero.  Next we will determine our period in terms of radians we can do so by looking at our period which happens to be 580.  This means that the line will repeat itself every 580 on the numberline and so:

P=\frac{2\pi }{B}\\ \\580=\frac{2\pi}{B}\\\\B=\frac{2\pi}{580} \\\\B=\frac{\pi}{290}

Therefore our period in radians is pi/290.

A sine function is such that:

y=Asin(Bx)+D

where A is the amplitude, B is the period, and D is the midline.  Our amplitude is 1, midline is zero and our period is pi/290 and so our final equation is:

y=sin(\frac{\pi}{290}x)


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c

Step-by-step explanation:

A is True

B is True

D is true


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3 years ago
I need to know the steps
vaieri [72.5K]

YES. She is correct.

\dfrac{x}{16}=4\qquad\text{multiply both sides by 4}\\\\\not\!4^1\cdot\dfrac{x}{16\!\!\!\!\!\diagup_4}=4\cdot4\\\\\dfrac{x}{4}=16

Equations are equivalent

Their solutions:

\dfrac{x}{16}=4\qquad\text{multiply both sides by 16}\\\\x=4\cdot16\\\\\boxed{x=64}\\----------\\\\\dfrac{x}{4}=16\qquad\text{multiply both sides by 4}\\\\x=16\cdot4\\\\\boxed{x=64}


3 0
3 years ago
A lamina occupies the part of the disk x2 + y2 ≤ 16 in the first quadrant. Find the center of mass of the lamina if the density
alekssr [168]

Answer: The center of mass =

(8192/(1280π), 8192/(1280π)).

Step-by-step explanation:

Here, δ(x,y) =k(x^2 + y^2) for some constant k.

So, m = ∫∫ δ(x,y) dA

..........= ∫(θ = 0 to π/2) ∫(r = 0 to 4) kr^2 × (r dr dθ), via polar coordinates

..........= (π/2) × (k/4)r^4 {for r = 0 to 4}

..........= 256πk/8.

My = ∫∫ x δ(x,y) dA

......= ∫(θ = 0 to π/2) ∫(r = 0 to 4) (r cos θ) × kr^2 × (r dr dθ), via polar coordinates

......= ∫(θ = 0 to π/2) cos θ dθ × ∫(r = 0 to 4) kr^4 dr 

......= sin θ {for θ = 0 to π/2} × (1/5)kr^5 {for r = 0 to 4}

......= 1024k/5.

Mx = ∫∫ y δ(x,y) dA

......= ∫(θ = 0 to π/2) ∫(r = 0 to 4) (r sin θ) × kr^2 × (r dr dθ), via polar coordinates

......= ∫(θ = 0 to π/2) sin θ dθ × ∫(r = 0 to 4) kr^4 dr 

......= -cos θ {for θ = 0 to π/2} × (1/5)kr^5 {for r = 0 to 4}

......= 1024k/5.

Hence, the center of mass is (My/m, Mx/m) =

My/m = 1024/5 ×8/256

The same for Mx/m the density at any point is proportional to the square of its distance from the origin

(8192/(1280π), 8192/(1280π)).

6 0
3 years ago
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