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Minchanka [31]
2 years ago
12

Xavier has $23.50 in his savings account. He deposits $35 every week. His mother also deposits $20 into the account every time X

avier washes the car. His savings account balance can be shown with the following expression:
20c + 35w + 23.50

Part A: Identify a coefficient, a variable, and a constant in this expression. (3 points)

Part B: If Xavier saves for 12 weeks and washes the car 3 times, how much will he have in his account? Show your work to receive full credit. (4 points)

Part C: If Xavier’s mother deposited $25 after each car wash, would the coefficient, variable, or constant in the expression change? Why? (3 points)
Mathematics
1 answer:
andrezito [222]2 years ago
8 0
Part A the coefficient is 20c and 35w, the variable is c and w representing cost and week. and the constant is 23.50.

Part B He would have 503.50 because <span><span>35</span><span>(12)</span></span>+60<span>=<span><span>420x</span>+<span>60+23.50 which would be 503.50

Part C The coefficient of the cost would change because you would be adding a 5 dollar increase!!!!!! done</span></span></span>
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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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