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sukhopar [10]
3 years ago
9

3[x]+9 what is the graph?

Mathematics
1 answer:
ipn [44]3 years ago
4 0
So, you can do this one of two ways.

The first is knowing what the numbers in your equation represent:

For example: y=mx+b in this case "b" is 9. b is the y-value when x=0. So the first point on our graph is (0,9). Next we have to pick an xvalue to solve for another y-coordinate.

I chose x= -3. Plug x into the equation to get y.
y=3x+9
y=3(-3)+9
y=(-9)+9
y=0

So, our second point is (-3,0).

connect the points with a ruler to graph the line.

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If a = {6 6 4 7 -4 4 -6 9 -3} and B= {1 5 -8 9 -5 -8 9 -5 2}, find 3a +9B Matrix operations
V125BC [204]

Answer:

3A+9B=\begin{bmatrix} 27 & 63 & -60 \\\\ 102 & -57 & -60\\\\ 63 & -18 & 9 \end{bmatrix}

Step-by-step explanation

  • A=\begin{bmatrix} 6 & 6 & 4\\\\7 & -4 & 4\\\\ -6 & 9 & -3\end{bmatrix}\: and \: B=\begin{bmatrix}1 & 5 & -8\\\\ 9 & -5 & -8\\\\ 9 & -5 & 2\end{bmatrix}

  • \rightarrow 3A=3\begin{bmatrix} 6 & 6 & 4\\\\7 & -4 & 4\\\\ -6 & 9 & -3\end{bmatrix},\:  \: 9B =9\begin{bmatrix}1 & 5 & -8\\\\ 9 & -5 & -8\\\\ 9 & -5 & 2\end{bmatrix}

  • \rightarrow 3A=\begin{bmatrix} 3*6 & 3*6 & 3*4\\\\3*7 & 3(-4) & 3*4\\\\ 3(-6) & 3*9 & 3(-3)\end{bmatrix},\:\: 9B =\begin{bmatrix}9*1 &9* 5 &9( -8)\\\\ 9*9 & 9(-5) & 9(-8)\\\\ 9*9 & 9(-5) & 9*2\end{bmatrix}

  • \rightarrow 3A=\begin{bmatrix} 18 & 18 & 12 \\\\ 21 & -12 & 12\\\\ -18 & 27 & -9\end{bmatrix},\:\: 9B =\begin{bmatrix} 9 & 45 & -72 \\\\ 81 & -45 & -72\\\\ 81 & -45 & 18\end{bmatrix}

  • \rightarrow 3A+9B=\begin{bmatrix}18+9 & 18+45 & 12+(-72) \\\\ 21+81 & -12 +(-45) & 12+(-72)\\\\ -18+81 & 27+(-45) & -9+18 \end{bmatrix}

  • \rightarrow 3A+9B=\begin{bmatrix} 18+9 & 18+45 & 12-72 \\\\ 21+81 & -12 -45 & 12-72\\\\ -18+81 & 27-45 & -9+18 \end{bmatrix}

  • \rightarrow \purple{\bold{3A+9B=\begin{bmatrix} 27 & 63 & -60 \\\\ 102 & -57 & -60\\\\ 63 & -18 & 9 \end{bmatrix}}}
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