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Maru [420]
3 years ago
14

Find the rectangular coordinates of the point with the polar coordinates ordered pair 7 comma 2 pi divided by 3.

Mathematics
1 answer:
Morgarella [4.7K]3 years ago
3 0

Answer:

\left(-\dfrac{7}{2},\dfrac{7\sqrt{3}}{2}\right).

Step-by-step explanation:

The given point is

\left(7,\dfrac{2\pi}{3}\right)

We need to find the rectangular coordinates of the given point.

If a polar coordinate is (r,\theta), then  

x=r\cos theta

y=r\sin theta

In the given point \left(7,\dfrac{2\pi}{3}\right),

r=7,\theta=\dfrac{2\pi}{3}

Now,

x=7\cos \dfrac{2\pi}{3}

x=7\cos \left(\pi-\dfrac{\pi}{3}\right)

x=-7\cos \left(\dfrac{\pi}{3}\right)

x=-7\left(\dfrac{1}{2}\right)

x=-\dfrac{7}{2}

and,

y=7\sin \dfrac{2\pi}{3}

y=7\sin \left(\pi-\dfrac{\pi}{3}\right)

y=7\sin \left(\dfrac{\pi}{3}\right)

y=7\left(\dfrac{\sqrt{3}}{2}\right)

y=\dfrac{7\sqrt{3}}{2}

Therefore, the required point is \left(-\dfrac{7}{2},\dfrac{7\sqrt{3}}{2}\right).

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To learn more about the rational root theorem, you can take a look at brainly.com/question/10937559

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Approximate the change in the volume of a sphere when its radius changes from r​ = 40 ft to r equals 40.05 ft (Upper V (r )equal
alexgriva [62]

Answer:

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

Step-by-step explanation:

The volume of the sphere (V), measured in cubic feet, is represented by the following formula:

V = \frac{4\pi}{3}\cdot r^{3}

Where r is the radius of the sphere, measured in feet.

The change in volume is obtained by means of definition of total difference:

\Delta V = \frac{\partial V}{\partial r}\Delta r

The derivative of the volume as a function of radius is:

\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}

Then, the change in volume is expanded:

\Delta V = 4\pi \cdot r^{2}\cdot \Delta r

If r = 40\,ft and \Delta r = 40\,ft-40.05\,ft = 0.05\,ft, the change in the volume of the sphere is approximately:

\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)

\Delta V \approx 1005.310\,ft^{3}

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

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