The equation x2 - 9 = 0 has real solution(s). DONE

Mathematics

1 answer:

Andru [333]2 years ago

5 0

Answer:

{-3, 3}

Step-by-step explanation:

x^2 - 9 = 0 can be written in factored form as (x -3)(x + 3) = 0. Letting each factor equal zero separately and solving for x, we get x: {-3, 3}

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Find (if possible) the complement and the supplement of each angle. (if not possible, enter impossible.) (a) π 10 complement rad

blagie [28]

In this question, we have to find the complement and supplement of the given angles .

Complementary angles are those angles whose sum is 90 degree and supplementary angles are those angles whose sum is 180 degree.

So to find the complement and supplement angles, we need to subtract the given angles from pi/2 and pi respectively .

$\frac{ \pi}{10}
\\
complement \ angle = \frac{ \pi}{2} - \frac{ \pi}{10} \\ = \frac{4 \pi}{10} = \frac{ 2 \pi}{5}
\\
Supplement \ angle = \pi - \frac{ \pi}{10} = \frac{9 \pi}{10}$

$\frac{9 \pi}{10}
\\
Complement \ angle = \frac{ \pi}{2} - \frac{9 \pi}{10} = \frac{-4 \pi}{10} = -\frac{2 \pi}{5}
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4 0

3 years ago

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 $\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\1\\0\end{array}\right]$