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stealth61 [152]
3 years ago
11

Which are coplanar and noncolinar?

Mathematics
1 answer:
Paul [167]3 years ago
3 0

<em>Coplanar</em> means "lying on the same plane," while <em>colinear </em>means "lying on the same line." In this problem, all four points A, B, C, and D are coplanar - they all lie on the plane \mathcal{H} - so we just need to find the points that lie on the same line. Though points C and B <em>would </em>be colinear if there was a line drawn between them, there's no such line in this problem, so we can rule that out. We notice one line drawn on the plane, and points A, C, and D lying on it, so we can say that points A, C, and D are both coplanar and colinear.

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Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify
irina [24]

Answer:

x-coordinates of relative extrema = \frac{-1}{2}

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}

Differentiate with respect to x

f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}

f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}

Differentiate f'(x) with respect to x

f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1

At x = \frac{-1}{2},

f''\left ( \frac{-1}{2} \right )=\frac{4\left ( -1+4\left ( \frac{-1}{2} \right ) \right )}{3\left ( \frac{-1}{2} \right )^{\frac{5}{3}}}>0

We know that if f''(a)>0 then x = a is a point of minima.

So, x=\frac{-1}{2} is a point of minima.

For inflexion points:

Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.

So, x-coordinates of the inflexion points are 0, 1

7 0
3 years ago
What is the correct reason for statement 3?
olchik [2.2K]
The answer would be Associative Property.

Hope this helps you =)
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torisob [31]

Answer:

6045000000000000000000000 kg.

Step-by-step explanation:

We have been given that the mass of Earth is 5.972\times10^{24} kg. The mass of the Moon is 7.3\times10^{22} kg.

To find the total mass we will add mass of Earth and Moon.

First of all let us convert the given masses in standard form.

5.972\times10^{24}=5972000000000000000000000  

7.3\times10^{22}=73000000000000000000000

5.972\times10^{24}+7.3\times10^{22}  

5972000000000000000000000+73000000000000000000000

6045000000000000000000000

Therefore, the mass of Earth and Moon is 6045000000000000000000000 kg.

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Answer:

option A

Step-by-step explanation:

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