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Nadusha1986 [10]
3 years ago
12

25 POINTS AVAILABLE

Mathematics
1 answer:
myrzilka [38]3 years ago
7 0

Answer:

\large\boxed{1.\ (-3, 0),\ r = 3}\\\boxed{2.\ (x+4)^2+(y-3)^2=36}\\\boxed{3.\ (x-2)^2+(y-1)^2=(\sqrt{34})^2}

Step-by-step explanation:

The equation of a circle in standard form:

(x-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

1. We have the equation:

(x+3)^2+y^2=9\\\\(x-(-3))^2+(y-0)^2=3^2

<h2 />

2. We have the center (-4, 3) and the radius r = 6. Substitute:

(x-(-4))^2+(y-3)^2=6^2\\\\(x+4)^2+(y-3)^2=36

3. We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)

Substitute:

h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Substitute the coordinates of the points (2, 1) and (5, -4):

r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}

Finally we have:

(x-2)^2+(y-1)^2=(\sqrt{34})^2

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Find all unit vectors that are orthogonal to the vector u = 1, 0, −4 .
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Answer:

Step-by-step explanation:

Given:

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If v = v₁ i + v₂ j + v₃ k is one of those vectors that are orthogonal to u, then

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Plug in the value of v₁ = 4v₃ into vector v as follows

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Equation (i) is the generalized form of all vectors that will be orthogonal to vector u

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Get the generalized unit vector by dividing the equation (i) by the magnitude of the generalized vector form. i.e

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where v₂ and v₃ are non-zero arbitrary real numbers.

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Step-by-step explanation:

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