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VikaD [51]
3 years ago
12

Find the rectangular coordinates of the point with the polar coordinates (8, 3 divided by 2 pi ). (1 point) (0, -8) (0, 8) (8, 0

) (-8, 0)
Mathematics
1 answer:
jolli1 [7]3 years ago
7 0

Answer:

(x,y) = (0,-8)

Step-by-step explanation:

We know that a point in polar coordinates is represented by

(r, θ)

Where r is the distance from the origin and θ is the angle.

Rectangular coordinates can be found by

x = r*cos(θ)

y = r*sin (θ)

x = r*cos(θ) = 8* cos (3π/2)

y = r*sin (θ) = 8 sin(3π/2)

x = 8* cos (3π/2) = 8*0 = 0

y = 8* sin (3π/2) = 8*(-1) = -8

(x,y) = (0,-8)

You might be interested in
A. Find the amplitude.
Feliz [49]

Answers:

  • a) Amplitude = 2
  • b) Period = pi
  • c) Vertical shift = -2, which means it has been shifted down 2 units.
  • d) Horizontal shift = 3pi/8, this shifting is to the right.
  • e) There is <u>  one  </u> cycle between 0 and 2pi.
  • f) The equation of the graph is y = 2*sin(2(x-3pi/8))-2

========================================================

Explanations:

Part (a)

The highest point is when y = 0 and the lowest point is when y = -4. The vertical distance between the peak and valley is 4 units, which cuts in half to 2. This is the amplitude. It's the vertical distance from the center to either the peak or valley.

Note: Amplitude is always positive as it measures a distance.

---------------------

Part (b)

For x > 0, the first valley or lowest point occurs between 0 and pi/4. It appears to be the midpoint of the two values. So that would be (0+pi/4)/2 = pi/8.

The next valley occurs between pi and 5pi/4. Compute the midpoint to get (pi+5pi/4)/2 = (4pi/4+5pi/4)/2 = (9pi/4)/2 = 9pi/8

So we have the graph go from one valley x = pi/8 to the next valley over x = 9pi/8. This is a distance of pi units because 9pi/8-pi/8 = 8pi/8 = pi

The graph repeats itself every pi units, so the period is pi.

---------------------

Part (c)

The midline is normally through y = 0, aka the x axis. However, the graph shows the midline is through y = -2. This means the graph has been shifted down 2 units.

---------------------

Part (d)

This will depend on whether you use sine or cosine. This is entirely because cosine is a phase-shifted version of sine, and vice versa. I'll go with sine.

The parent sine function y = sin(x) goes through the origin (0,0)

However, as part (c) mentioned, we shifted the graph 2 units down. So we have y = sin(x)-2. But plugging x = 0 into this leads to the point (0,-2)

This doesn't match what the graph says. The graph shows the point (3pi/8, -2) on the red curve. The x coordinate 3pi/8 is the midpoint of pi/4 and pi/2

This must mean we need to shift the sine graph 3pi/8 units to the right.

---------------------

Part (e)

Start at the lowest point when x = pi/8. If you start the cycle here, then it ends when x = 9pi/8. See part (b).

So far we've completed one cycle. If we start at x = 9pi/8, then the next valley or lowest point is slightly beyond or to the right of x = 2pi. This means we run out of room and we haven't completed a full cycle.

Overall, one full cycle is between 0 and 2pi.

---------------------

Part (f)

Again I'm going to use sine instead of cosine. Refer back to part (d).

The general sine curve equation is

y = A*sin(B(x-C))+D

where

  • |A| = amplitude
  • B handles the period, specifically T = 2pi/B where T is the period. We can solve for B to get B = 2pi/T
  • C = horizontal phase shift
  • D = vertical shift, and ties together with the midline equation

In this case, we found that

  • |A| = 2
  • T = pi leads to B = 2pi/T = 2pi/pi = 2
  • C = 3pi/8
  • D = -2

So,

y = A*sin(B(x-C))+D

will update to

y = 2*sin(2(x-3pi/8))-2

which is one way to express the equation of the red curve. Optionally you can distribute the 2 through to (x-3pi/8).

6 0
2 years ago
Please help me divide
Inessa [10]
Hope this work helps you

3 0
3 years ago
Read 2 more answers
Which figures are polygons
Tatiana [17]
The answer is A. A, D the reason being, is because polygons have sides C, and B have no sides so we can eliminate those immediately.
5 0
3 years ago
The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E
nexus9112 [7]

Answer:

The projected enrollment is \lim_{t \to \infty} E(t)=10,000

Step-by-step explanation:

Consider the provided projected rate.

E'(t) = 12000(t + 9)^{\frac{-3}{2}}

Integrate the above function.

E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt

E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c

2000=-\frac{24000}{3}+c

2000=-8000+c

c=10,000

Therefore, E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

Now we need to find \lim_{t \to \infty} E(t)

\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

\lim_{t \to \infty} E(t)=10,000

Hence, the projected enrollment is \lim_{t \to \infty} E(t)=10,000

8 0
3 years ago
PLEASE HELP ME me me me​
kykrilka [37]

Answer:

Andre has the correct answer. When simplified his answer is equivalent to the original equation.

7 0
3 years ago
Read 2 more answers
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