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3 years ago

Find the equation of the circle with center at (-1, 3) and radius of 4.

Mathematics

1 answer:

Nataly [62]3 years ago

7 0

Answer:

(x+1)^2 + (y-3)^2 = 16

Step-by-step explanation:

The equation of a circle is given by

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

Where (h,k) is the center and r is the radius

(x--1)^2 + (y-3)^2 = 4^2

(x+1)^2 + (y-3)^2 = 4^2

(x+1)^2 + (y-3)^2 = 16

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16. Let W be the set of all vectors in R3 of the form a+2b b -3a Find a basis for Wand state the dimension of W.

Yuki888 [10]

Answer:

$W=\{\left[\begin{array}{ccc}a+2b\\b\\-3a\end{array}\right]: a,b\in\mathbb{R} \}$

Observe that if the vector $x=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ is in W then it satisfies:

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{c}a+2b\\b\\-3a\end{array}\right]=a\left[\begin{array}{c}1\\0\\-3\end{array}\right]+b\left[\begin{array}{c}2\\1\\0\end{array}\right]$

This means that each vector in W can be expressed as a linear combination of the vectors $\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]$

Also we can see that those vectors are linear independent. Then the set

$\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]\}$ is a basis for W and the dimension of W is 2.

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3 years ago