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victus00 [196]
3 years ago
9

Find -3A+6B Matrices

Mathematics
1 answer:
natima [27]3 years ago
6 0
\bf A=
\begin{bmatrix}
-3&5&-6\\9&-5&3
\end{bmatrix}\qquad \qquad 
B=
\begin{bmatrix}
-2&6&7\\2&-1&-6
\end{bmatrix}\\\\
-------------------------------\\\\
-3A=
\begin{bmatrix}
9&-15&18\\-27&15&-9
\end{bmatrix}\qquad \qquad 6B=
\begin{bmatrix}
-12&36&42\\12&-6&-36
\end{bmatrix}\\\\
-------------------------------\\\\
-3A+6B\implies 
\begin{bmatrix}
9-12&\qquad -15+36&\qquad 18+42\\
-27+12&\qquad 15-6&\qquad -9-36
\end{bmatrix}
\\\\\\
-3A+6B\implies 
\begin{bmatrix}
-3&21&60\\
-15&9&-45
\end{bmatrix}
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2 years ago
\angle a∠aangle, a and \angle b∠bangle, b are complementary angles. \angle a∠aangle, a measures 64^\circ64
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The width of a rectangle measures (3s+t) centimeters,and it’s length measures (3s-9t) centimeters. Which expression represents t
andreyandreev [35.5K]

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Step-by-step explanation:

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3 0
3 years ago
) Picture frames are produced such that the four sides of a picture frame consist of two pieces from a population whose mean len
Natali [406]

Answer:

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Step-by-step explanation:

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Perimeter of rectangle = 2(A + B)

Going by the illustration in the question, the perimeter is

==> 2(32 + 44) = 152 cm.

However, we a given the mean perimeter of the frame which are:

32 and 44. We can either choose to say; each sides is (32 by 32) and (44 by 44). Or we consider it in mean/average term as given: (32 by 44).

Case 1: To obtain the mean perimeter of the frame using the first scenario, we say:

Mean perimeter = (32+32+44+44)/4 = 38 cm.

Case 2: We can simply say:

Mean perimeter = (32 + 44)/2 = 38 cm.

Graphically: Assuming that the frame is given below

                     32 cm

          ---------------------------------

         |                                      |

         |                                      |

44 cm|                                      | 44 cm

         |                                      |

         |                                      |

         |                                      |

         ----------------------------------

                       32 cm

The mean of upper frame/bars = 32 cm

The mean of side frame/bars  =  44 cm

Hence, the mean of the perimeter = (32 + 44)/2 = 38 cm

OR,

we take the sum of all the sides (32 + 44 + 32 + 44) and divide by total number of sides (4). That is;

The mean of the perimeter = (32 + 44 + 32 + 44)/4 = 38 cm

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AURORKA [14]
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