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irina1246 [14]
4 years ago
13

What is x^3=8/27? pls help

Mathematics
1 answer:
yaroslaw [1]4 years ago
4 0

Answer:x=2/3 =0.6667

Step-by-step explanation:

...

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Find the points on x^2+3(y-2)^2 = 27<br> which are closest to and farthest from the origin
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(-4.89, 3) and (4.89, 3) are the <em>farthest</em> from the origin and (0, -1) is the <em>closest</em> from the origin.

Let be a curve of the form x^{2}+3\cdot (y-2)^{2} = 27. The distance with respect to origin is found by using the following Pythagorean identity:

r = x^{2}+y^{2} (1)

Where r is the square distance function.

We can modify (1) as following:

r = [27-3\cdot (y-2)^{2}]+y^{2}

r = 27-3\cdot (y^{2}-4\cdot y +4)+y^{2}

r = 27-3\cdot y^{2}+12\cdot y -12 + y^{2}

r = -2\cdot y^{2}+12\cdot y+27 (1b)

Now we apply the First and second derivative tests to determine the minimum and maximum distances from the origin:

First derivative test

-4\cdot y +12 = 0

y = 3

Second derivative test

r'' = -4

The y-component represents a maximum.

Now we graph the function with the ressource of a graphing tool, we find the following points:

Farthest points: (-4.89, 3), (4.89, 3).

Closest points: (0, -1).

(-4.89, 3) and (4.89, 3) are the <em>farthest</em> from the origin and (0, -1) is the <em>closest</em> from the origin.

To learn more on ellipses, we kindly invite to check this verified question: brainly.com/question/19507943

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melomori [17]

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Step-by-step explanation:

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