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bazaltina [42]
3 years ago
11

Enter the equations of the asymptotes for the function f(x). f(x)= 3/x−7 + 2

Mathematics
1 answer:
Alexxx [7]3 years ago
3 0

Answer:

x = 7, y = 2

Step-by-step explanation:

I assume it is 3/(x-7) + 2.

When x = 7, there is an asymptote because it is undefined.

When y = 2, there is also one, because 3/(x-7) is never 0.

These are the only ones.

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7-0.3×&gt;4<br>solve for x​
Troyanec [42]

Answer:

x<10

Step-by-step explanation:

Add '-7' to each side of the equation.

7 + -7 + -0.3x = 4 + -7

Combine like terms: 7 + -7 = 0

0 + -0.3x = 4 + -7

-0.3x = 4 + -7

Combine like terms: 4 + -7 = -3

-0.3x = -3

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3 years ago
Which relationship is always true for the angles RXY and Z or the triangle ABC
aksik [14]

Answer:

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x + y + z = 180 degrees

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Step-by-step explanation:

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3 years ago
What eigen value for this matix <br> (1 -2)<br> (-2 0)
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)
  2. Compute the determinant of A'
  3. Set the determinant of A' equal to zero and solve for lambda.

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}

5 0
3 years ago
Read 2 more answers
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