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xenn [34]
3 years ago
12

M=x^3+y^2-6(x-y)-2021

Mathematics
2 answers:
EastWind [94]3 years ago
8 0

Answer:

Not sure explain it again

Step-by-step explanation:

B

Harrizon [31]3 years ago
5 0
M=x^3+y^3-6x+6y-2021
=x^3-6x+y^3+6y-2021
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5.4

Step-by-step explanation:

I am not really sure about this type of problem, but a diameter has to cross the midpoint of the circle the dark dot and hit two point on both side of the circle. Yet the line 9.6 is not a diameter since it didn't go through the center, so the diameter has to be longer than 9.6. But x is looking for radius. so divide 9.6 by 2 u get 4.8 but that is not a diameter and the real diameter is longer that mean the radius is longer so the only answer is 5.4

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Step-by-step explanation:

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Find the angle between u =the square root of 5i-8j and v =the square root of 5i+j.
fenix001 [56]

Answer:

The angle between vector \vec{u} = 5\, \vec{i} - 8\, \vec{j} and \vec{v} = 5\, \vec{i} + \, \vec{j} is approximately 1.21 radians, which is equivalent to approximately 69.3^\circ.

Step-by-step explanation:

The angle between two vectors can be found from the ratio between:

  • their dot products, and
  • the product of their lengths.

To be precise, if \theta denotes the angle between \vec{u} and \vec{v} (assume that 0^\circ \le \theta < 180^\circ or equivalently 0 \le \theta < \pi,) then:

\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}.

<h3>Dot product of the two vectors</h3>

The first component of \vec{u} is 5 and the first component of \vec{v} is also

The second component of \vec{u} is (-8) while the second component of \vec{v} is 1. The product of these two second components is (-8) \times 1= (-8).

The dot product of \vec{u} and \vec{v} will thus be:

\begin{aligned} \vec{u} \cdot \vec{v} = 5 \times 5 + (-8) \times1 = 17 \end{aligned}.

<h3>Lengths of the two vectors</h3>

Apply the Pythagorean Theorem to both \vec{u} and \vec{v}:

  • \| u \| = \sqrt{5^2 + (-8)^2} = \sqrt{89}.
  • \| v \| = \sqrt{5^2 + 1^2} = \sqrt{26}.

<h3>Angle between the two vectors</h3>

Let \theta represent the angle between \vec{u} and \vec{v}. Apply the formula\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} to find the cosine of this angle:

\begin{aligned} \cos(\theta)&= \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} = \frac{17}{\sqrt{89}\cdot \sqrt{26}}\end{aligned}.

Since \theta is the angle between two vectors, its value should be between 0\; \rm radians and \pi \; \rm radians (0^\circ and 180^\circ.) That is: 0 \le \theta < \pi and 0^\circ \le \theta < 180^\circ. Apply the arccosine function (the inverse of the cosine function) to find the value of \theta:

\displaystyle \cos^{-1}\left(\frac{17}{\sqrt{89}\cdot \sqrt{26}}\right) \approx 1.21 \;\rm radians \approx 69.3^\circ .

3 0
3 years ago
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