Answer:
D.) You can show your name printed on the team roster.Step-by-step explanation:
it would be known as a mixed fraction....?
$15.75
2 x 7.5 = 15
0.10 x 7.5 = 0.75
15 + 0.75 = $15.75
Hope this helped
Each element of the matrix are multiplied by the scalar to form a matrix of
same size as the original matrix in matrix scalar multiplication.
Reasons:
The matrix <em>A</em> is presented as follows;
![A = {\left[\begin{array}{ccc}4&6&8\\6&8&10\end{array}\right]}](https://tex.z-dn.net/?f=A%20%3D%20%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%266%268%5C%5C6%268%2610%5Cend%7Barray%7D%5Cright%5D%7D)
Using the multiplication of a matrix and a scalar, we have;
![60 \cdot A = 60 \cdot \left[\begin{array}{ccc}4&6&8\\6&8&10\end{array}\right] = \left[\begin{array}{ccc}60 \times 4&60 \times 6&60 \times 8\\60 \times 6&60 \times 8&60 \times 10\end{array}\right] = \left[\begin{array}{ccc}\mathbf{240}&\mathbf{360}&\mathbf{480}\\\mathbf{360}&\mathbf{480}&\mathbf{600}\end{array}\right]](https://tex.z-dn.net/?f=60%20%5Ccdot%20A%20%3D%2060%20%5Ccdot%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%266%268%5C%5C6%268%2610%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D60%20%5Ctimes%204%2660%20%5Ctimes%206%2660%20%5Ctimes%208%5C%5C60%20%5Ctimes%206%2660%20%5Ctimes%208%2660%20%5Ctimes%2010%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cmathbf%7B240%7D%26%5Cmathbf%7B360%7D%26%5Cmathbf%7B480%7D%5C%5C%5Cmathbf%7B360%7D%26%5Cmathbf%7B480%7D%26%5Cmathbf%7B600%7D%5Cend%7Barray%7D%5Cright%5D)
Therefore;
![60 \cdot A = \left[\begin{array}{ccc}240\4&360&480\\360&480&600\end{array}\right]](https://tex.z-dn.net/?f=60%20%5Ccdot%20A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D240%5C4%26360%26480%5C%5C360%26480%26600%5Cend%7Barray%7D%5Cright%5D)
Learn more about matrices here:
brainly.com/question/14296012
C: none of these are solutions to the given equation.
• If<em> y(x)</em> = <em>e</em>², then <em>y</em> is constant and <em>y'</em> = 0. Then <em>y'</em> - <em>y</em> = -<em>e</em>² ≠ 0.
• If <em>y(x)</em> = <em>x</em>, then <em>y'</em> = 1, but <em>y'</em> - <em>y</em> = 1 - <em>x</em> ≠ 0.
The actual solution is easy to find, since this equation is separable.
<em>y'</em> - <em>y</em> = 0
d<em>y</em>/d<em>x</em> = <em>y</em>
d<em>y</em>/<em>y</em> = d<em>x</em>
∫ d<em>y</em>/<em>y</em> = ∫ d<em>x</em>
ln|<em>y</em>| = <em>x</em> + <em>C</em>
<em>y</em> = exp(<em>x</em> + <em>C </em>)
<em>y</em> = <em>C</em> exp(<em>x</em>) = <em>C</em> <em>eˣ</em>
