Question

Find the values of x and y from the matrix relation:

$\begin{gathered} \left[\begin{array}{ccc}3&-2\\-1&4\end{array}\right] \times\left[\begin{array}{ccc}2x\\1\end{array}\right] + 2 \left[\begin{array}{ccc}-4\\5\end{array}\right] = \left[\begin{array}{ccc}8\\4y\end{array}\right] \end{gathered}$​

Answer 1

Answer:

Answer

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∫x(x6+1)1dx

=∫x(x6+1)x6+1−x6dx

=∫x(x6+1)x6+1dx−∫x(x6+1)x6dx

=∫xdx−∫(x6+1)x5dx

Let t=x6+1⇒dt=6x5dx

=∫xdx−61∫tdt

=logx−61log∣t∣+c

thanks

Answer 2

Answer:

• Step 1: List all the given. points. • Klara answered of the test correctly. we use Step 2: Determine the problem and the operation to use. Since Klara got of the to determine her score. To evaluate numbers with different form, you can choose either of the following: • Convert all numbers to decimal. . Convert all numbers to fraction. Step 3: Evaluate and proceed with the necessary operation. Answer:

$\begin{gathered} \left[\begin{array}{ccc}3&-2\\-1&4\end{array}\right] \times\left[\begin{array}{ccc}2x\\1\end{array}\right] + 2 \left[\begin{array}{ccc}-4\\5\end{array}\right] = \left[\begin{array}{ccc}8\\4y\end{array}\right] \end{gathered}$

$\begin{gathered} \left[\begin{array}{ccc}3&-2\\-1&4\end{array}\right] \times\left[\begin{array}{ccc}2x\\1\end{array}\right] + 2 \left[\begin{array}{ccc}-4\\5\end{array}\right] = \left[\begin{array}{ccc}8\\4y\end{array}\right] \end{gathered}$

$\begin{gathered} \left[\begin{array}{ccc}3&-2\\-1&4\end{array}\right] \times\left[\begin{array}{ccc}2x\\1\end{array}\right] + 2 \left[\begin{array}{ccc}-4\\5\end{array}\right] = \left[\begin{array}{ccc}8\\4y\end{array}\right] \end{gathered}$

$\begin{gathered} \left[\begin{array}{ccc}3&-2\\-1&4\end{array}\right] \times\left[\begin{array}{ccc}2x\\1\end{array}\right] + 2 \left[\begin{array}{ccc}-4\\5\end{array}\right] = \left[\begin{array}{ccc}8\\4y\end{array}\right] \end{gathered}$