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Mkey [24]
3 years ago
7

What is an equation of the line that passes through the point (-4,-3) and is perpendicular to the line 2x+3y=15

Mathematics
1 answer:
Katen [24]3 years ago
5 0

Answer:

The equation would be y = 3/2x + 3

Step-by-step explanation:

First we have to find the slope of the original line. We can do this by solving for y.

2x + 3y = 15

3y = -2x + 15

y = -2/3x + 5

Now that we have the slope of -2/3, we know the slope of the new line will be 3/2. This is because perpendicular lines have opposite and reciprocal slopes. We can use that and a point in point-slope form to find the equation.

y - y1 = m(x - x1)

y + 3 = 3/2(x + 4)

y + 3 = 3/2x + 6

y = 3/2x + 3

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Answer:

A= 1/8
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Solution:
Count how many lines there are except the Line on zero. This number is 8 and will be the denominator. Then count the number of lines to the one you’re looking at
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Find all of the eigenvalues λ of the matrix A. (Hint: Use the method of Example 4.5 of finding the solutions to the equation 0 =
Svetradugi [14.3K]

Answer:

\lambda=8,\ \lambda=-5

Step-by-step explanation:

<u>Eigenvalues of a Matrix</u>

Given a matrix A, the eigenvalues of A, called \lambda are scalars who comply with the relation:

det(A-\lambda I)=0

Where I is the identity matrix

I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]

The matrix is given as

A=\left[\begin{array}{cc}3&5\\8&0\end{array}\right]

Set up the equation to solve

det\left(\left[\begin{array}{cc}3&5\\8&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda \end{array}\right]\right)=0

Expanding the determinant

det\left(\left[\begin{array}{cc}3-\lambda&5\\8&-\lambda\end{array}\right]\right)=0

(3-\lambda)(-\lambda)-40=0

Operating Rearranging

\lambda^2-3\lambda-40=0

Factoring

(\lambda-8)(\lambda+5)=0

Solving, we have the eigenvalues

\boxed{\lambda=8,\ \lambda=-5}

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3 years ago
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42mph

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