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Solnce55 [7]
3 years ago
5

21a+9 Factor the algebraic expression.

Mathematics
1 answer:
Mama L [17]3 years ago
8 0
21a + 9 = 0 that is the answer good luck
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What is the minimum value for the function shown in the graph?<br><br> Enter your answer in the box.
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The minimum value is the lowest point of a graphed line. The would be the bottom of the "U" shape.

Looking at the graph you can see that the bottom of the U is touching the horizontal line at x = -5.5, so this would be the minimum value.

The lowest part of the line is at -5.5

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Find the area of the figure please
Eva8 [605]

Answer:

16m

Step-by-step explanation:

Separate the rectangle from the triangle. The area of the rectange is <u>10m</u>. We know this because <em>l x w</em> (2m x 5m).

Now for the triangle;

The area of a triangle is <em>b x h / 2. </em>We take the full height of the figure and subtract 5 from it giving us 3 which is our height for the triangle. We then see that 2m is on the bottom and we add that 2m to the two separate 1m giving us 4m in total.

Now that we have the base and height of the triangle we plug in giving us 6m. Finally we add both the areas; 10m + 6m = <u>16m</u>.

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natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

  1. Compute the matrix A' = A-\lambda I, where I is the identity matrix (1s on the diagonal, 0s elsewhere)
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So, in this case, we have

A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]

The determinant of this matrix is

\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4

Finally, we have

\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}

So, the two eigenvalues are

\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}

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Step-by-step explanation:

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