Question

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Answer 1

@Narad

Answer 2

Are you familiar with matrices and their rules?

Just multiply the outside term to all the numbers in the matrix. You will get option C.

Answer 3

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can you press on that @Narad ?

Answer 4

Add the corresponding positions:
[ 4+(-1) 6+0]
[ 7+9 1+2]


It's option A

Answer 5

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Answer 6

True

Answer 7

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Answer 8

Apply the rules of matrices:
[A C][X Z]
[B D][Y W]
=
[(AX+CY) (AZ+CW)]
[(BX+DY) (BZ+DW)]

Use this and I think you can solve it.

Answer 9

okay

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Answer 10

Consistent system = one solution only
dependently consistent solution = infinite solutions
Inconsistent = no solution.

The lines are parallel. Use this logic and you can find the answer.

Answer 11

so its false

Answer 12

Correct.

Answer 13

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true?

Answer 14

First compare the slopes of the two lines. What do you think they are?

Answer 15

It's false, m1=m2, where m1=-2 and m2=-2.
When slopes are the same, the lines are parallel, or they would never intersect. There's no solution, it should be false.

Answer 16

I mean -1* for both.

Answer 17

Ohh okay got it

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D?

Answer 18

yes

Answer 19

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Answer 20

plug in x as 2y in the first equation, then solve for y. once you get your y value find x with the second equation. Then you'll get your coordinates.

Answer 21

okay

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Answer 22

Same logic here as well. Plug in y as 3x-12 in the second equation, find x, then find the y value by plugging in the x value you obtained into the first equation.

Answer 23

okay

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Answer 24

convert both of the equations so that either the x or y coefficients are at their opposite values of the least common multiple (ie for 2y and 6y, the least common multiple is 6, so convert the first equation by multiplying the equation by 3, then second by multiplying by -1 so you can eliminate the y variable). Then find the x. After finding x, plug in the value and get y.

Answer 25

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Answer 26

Let one number be = x

The other number is = x+7

The sum of the numbers is

\[x+x+7=63\]
\[2x+7=63\]
\[2x=63-7=56\]
\[x=56/2=28\]
The numbers are 28 and 35
The answer is option C

Answer 27

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Answer 28

Question nº16
The equations are
\[3x-7y=25\] and\[5x-8y=27\]
The matrix equation is
\[\left[\begin{matrix}3 & -7\\ 5 & -8\end{matrix}\right] *\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}25 \\ 27\end{matrix}\right)\]
The inverse of matrix
\[A=\left[\begin{matrix}3 & -7\\ 5 & -8\end{matrix}\right]\]
is
\[A ^{-1}=1/detA*\left[\begin{matrix}-8 & 7 \\ -5& 3\end{matrix}\right]\]
\[=\left[\begin{matrix}-8/11 & 7/11\\ -5/11& 3/11\end{matrix}\right]\]
And the solution is \[\left(\begin{matrix}x \\ y\end{matrix}\right)=A ^{-1}*\left(\begin{matrix}25 \\ 27\end{matrix}\right)=\left(\begin{matrix}-1 \\ -4\end{matrix}\right)\]
The solution is option C

Answer 29

Question nº17
The matrix is
\[A=\left[\begin{matrix}7 & 5 \\ 4 & 3\end{matrix}\right]\]
The inverse is
\[A ^{-1}=\left[\begin{matrix}3 & -5\\ -4 & 7\end{matrix}\right]\]
The solution is \[\left(\begin{matrix}x \\ y\end{matrix}\right)=A ^{-1}*\left(\begin{matrix}14 \\ 9\end{matrix}\right)\]
\[=\left(\begin{matrix}-3 \\ 7\end{matrix}\right)\]
The answer is option B

Answer 30

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Answer 31

The first equation is
\[9x+2y=2\]
Multiply by 3\[27x+6y=6\]
Subtract the second equation
\[21x+6y=4\]
\[27x-21x=6x=6-4=2\]
\[x=2/6=1/3\] and
\[2y=2-9x=2-9*1/3=2-3=-1\]
\[y=-1/2\]
The solution is (1/3, -1/2)

Answer 32

Question nº15
The first eqaution is
\[2x-5y=1\]
Multiply by 4
\[8x-20y=4\]
The second equation is
\[3x-4y=-2\]
Multiply by 5
\[15x-20y=-10\]
Subtract the equations
\[8x-15x=7x=4-(-10)=14\]
\[x=2\] and
\[5y=-1-2x=-1-4=-5\]
\[y=-1\]
The solution is (-2,-1)

Answer 33

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Answer 34

This is done

Answer 35

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Answer 36

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Answer 37

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Answer 38

Question nº12

Substitute the equation
\[x=2y\]
in the first equation
\[4y+8y=12y=1\]
\[y=1/12\] and
\[x=2/12=1/6\]
The solutions are (1/6,1/12 )

Answer 39

Question nº13
Substitute the equation
\[y=3x-12\]
in the second equation
\[2x-3(3x-12)=-13\]
\[2x-9x+36=-13\]
\[-7x=-13-39=-49\]
\[x=-49/-7=7\] and
\[y=3*7-12=21-19=9\]
The solution is
{7,9}

Answer 40

okay

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Answer 41

Question nº 5
This is matrix multiplication
\[\left[\begin{matrix}4& -3 \\ 2 & 1\end{matrix}\right]*\left[\begin{matrix}1 & -2 \\ 2 & 1\end{matrix}\right]\]
\[=\left[\begin{matrix}-2 & -11\\ 4 & -3\end{matrix}\right]\]
The answer is option A

Answer 42

Question nº6
This is matrix multiplication
\[\left[\begin{matrix}4 & 5 \\ -2 & 1\\3&0\end{matrix}\right] *\left[\begin{matrix}1 & -2&1\\ 0 & 3&-4\end{matrix}\right]\]
=\[\left[\begin{matrix}4 & 7&-16\\ -2 & 7&-6\\3&-6&3\end{matrix}\right]\]
The answer is option B

Answer 43

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Answer 44

The first (2,-2) is not a solution to the system of inequalities since
\[x >-2\]
The answer is A

Answer 45

okay

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Answer 46

The equations are
\[y>-x+2\] and
\[y< 2x-1\]
The answer is option A

Answer 47

okay

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Answer 48

Question nº21
The augmented matrix is
A=\[\left[\begin{matrix}3 & -1&0&-1\\ 2& -1&1&-6\\1&4&-1&9\end{matrix}\right]\]
Perform row operations on the augmented matrix
The solution is (0,1,-5)
This is solution A

Answer 49

Question nº22
The augmented matrix is
\[\left[\begin{matrix}2 &-4 &3&-8 \\ 1 & 3&-2&9\\3&2&1&13\end{matrix}\right]\]
Perform row operations on the augmented matrix
The solution is (1,4,2)
The answer is option A

Answer 50

Question nº23
Let the numbers be x, y and z
The equations are
\[x+y+z=11\]
\[2y+z=x\]
\[x=2y\]
The augmented matrix is
\[\left[\begin{matrix}1 & 1&1&11 \\ -1 & 3&1&0\\1&-2&0&0\end{matrix}\right]\]
Perform row operations on this matrix
The solutions are (11, 11/2, -11/2)
This is option D

Answer 51

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Answer 52

Question nº24
This is a matrix subtraction
\[\left[\begin{matrix}0 & 2\\ 9 & -4\end{matrix}\right]-\left[\begin{matrix}-2 & 10\\ 3 & 5\end{matrix}\right]\]
\[=\left[\begin{matrix}2 & -8 \\ 6 & -9\end{matrix}\right]\]

Answer 53

Question nº25
This is a matrix addition
\[\left[\begin{matrix}0 & 2 \\ 9 & -4\end{matrix}\right] +\left[\begin{matrix}-2 & 10 \\ 3 & 5\end{matrix}\right]\]
\[=\left[\begin{matrix}-2 & 12 \\ 12 & 1\end{matrix}\right]\]