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bearhunter [10]

2 years ago

14

Identify the slope of the graphed line: Identify the y-intercept of the graphed line: Identify the slope of the line given by th

e equation: Identify the y-intercept of the line given by the equation:

2 answers:

aniked [119]2 years ago

7 0

PART 1: Identify the slope of the graphed line

To find the slope, I will use the slope formula: m = (y₂ - y₁) / (x₂ - x₁) using the points (0, 1) & (3, 0).

m = (0 - 1) / (3 - 0)

m = - 1 / 3

The slope of the graphed line is negative one-third: - 1/3.

PART 2: Identify the y-intercept of the graphed line

The y-intercept of a line is the point where the line crosses the y-axis. In this problem, the line crosses the y-axis at y = 1.

PART 3: Identify the slope of the line given by the equation

The given equation is written in the slope-intercept form: y = mx + b, where m is the slope and b is the y-intercept. The slope of the line as shown by the equation is 1/2.

PART 4: Identify the y-intercept of the line given by the equation

As previously stated, the equation is written is the slope-intercept form, so to find the y-intercept in the equation, all we need to do is find the value for b. In this case, b = -1.

Hope this helps!

olya-2409 [2.1K]2 years ago

7 0

Identify the slope of the graphed line:

✔ -1/3

Identify the y-intercept of the graphed line:

✔ 1

Identify the slope of the line given by the equation:

✔ 1/2

Identify the y-intercept of the line given by the equation:

✔ -1

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Find c1 and c2 such that M2+c1M+c2I2=0, where I2 is the identity 2×2 matrix and 0 is the zero matrix of appropriate dimension.

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The question is missing parts. Here is the complete question.

Let M = $\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]$ . Find $c_{1}$ and $c_{2}$ such that $M^{2}+c_{1}M+c_{2}I_{2}=0$ , where $I_{2}$ is the identity 2x2 matrix and 0 is the zero matrix of appropriate dimension.

Answer: $c_{1} = \frac{-16}{10}$

$c_{2}=\frac{-214}{10}$

Step-by-step explanation: Identity matrix is a sqaure matrix that has 1's along the main diagonal and 0 everywhere else. So, a 2x2 identity matrix is:

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

$M^{2} = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]$

$M^{2}=\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]$

Solving equation:

$\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+c_{1}\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right] +c_{2}\left[\begin{array}{cc}1&0\\0&1\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$

Multiplying a matrix and a scalar results in all the terms of the matrix multiplied by the scalar. You can only add matrices of the same dimensions.

So, the equation is:

$\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+\left[\begin{array}{cc}6c_{1}&5c_{1}\\-1c_{1}&-4c_{1}\end{array}\right] +\left[\begin{array}{cc}c_{2}&0\\0&c_{2}\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$

And the system of equations is:

$6c_{1}+c_{2} = -31\\-4c_{1}+c_{2} = -15$

There are several methods to solve this system. One of them is to multiply the second equation to -1 and add both equations:

$6c_{1}+c_{2} = -31\\(-1)*-4c_{1}+c_{2} = -15*(-1)$

$6c_{1}+c_{2} = -31\\4c_{1}-c_{2} = 15$

$10c_{1} = -16$

$c_{1} = \frac{-16}{10}$

With $c_{1}$ , substitute in one of the equations and find $c_{2}$ :

$6c_{1}+c_{2}=-31$

$c_{2}=-31-6(\frac{-16}{10} )$

$c_{2}=-31+(\frac{96}{10} )$

$c_{2}=\frac{-310+96}{10}$

$c_{2}=\frac{-214}{10}$

<u>For the equation, </u> $c_{1} = \frac{-16}{10}$ <u> and </u> $c_{2}=\frac{-214}{10}$ <u />

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