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Licemer1 [7]
3 years ago
8

How many solutions does the system have?

Mathematics
1 answer:
nikdorinn [45]3 years ago
5 0

Answer:

none

Step-by-step explanation:

there is no solutions to this system

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2 students are playing Rock, Paper, Scissors.
LUCKY_DIMON [66]

Answer:

1/3

Step-by-step explanation:

Since there are three option for Player A to choose which are

Rock , Paper , Scissors

The probability he choose paper is 1/3 , because

there are total three options so the denominator is the total number of outcomes which are 3, and the numerator is the outcome he choose which is paper it means he chooses one outcome so hence the probability of Player A choosing paper is 1/3

4 0
3 years ago
??????????????????????????
a_sh-v [17]

Answer:

Not a polynomial.

Step-by-step explanation:

A polynomial would have addition or subtraction signs separating the variable terms. so while -7c^3 d is not a polynomial, -7c^3 +cd would be.

5 0
3 years ago
When n is divided by 3, the remainder is 1 What is the remainder when 2n is divided by 3<br>​
MrMuchimi

Answer:

Only n if 2n is/3 when 2nis divide by 3

8 0
2 years ago
Read 2 more answers
9\7.80<br> 12\13.80<br> 34.11\9
Mazyrski [523]

Answer:

9/7.80=1.1538

12/13.80=0.8696

34.11/9=3.79

Step-by-step explanation:

8 0
3 years ago
If A = 1 2 1 1 and B= 0 -1 1 2 then show that (AB)^-1 = B^-1 A^-1<br><br><br> help meeeee plessss ​
Trava [24]

A = \begin{bmatrix}1&2\\1&1\end{bmatrix} \implies A^{-1} = \dfrac1{\det(A)}\begin{bmatrix}1&-1\\-2&1\end{bmatrix} = \begin{bmatrix}-1&1\\2&-1\end{bmatrix}

where det(<em>A</em>) = 1×1 - 2×1 = -1.

B = \begin{bmatrix}0&-1\\1&2\end{bmatrix} \implies B^{-1} = \dfrac1{\det(B)}\begin{bmatrix}2&1\\-1&0\end{bmatrix} = \begin{bmatrix}2&1\\-1&0\end{bmatrix}

where det(<em>B</em>) = 0×2 - (-1)×1 = 1. Then

B^{-1}A^{-1} = \begin{bmatrix}2&1\\-1&0\end{bmatrix} \begin{bmatrix}-1&1\\2&-1\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

On the other side, we have

AB = \begin{bmatrix}1&2\\1&1\end{bmatrix} \begin{bmatrix}0&-1\\1&2\end{bmatrix} = \begin{bmatrix}2&3\\1&1\end{bmatrix}

and det(<em>AB</em>) = det(<em>A</em>) det(<em>B</em>) = (-1)×1 = -1. So

(AB)^{-1} = \dfrac1{\det(AB)}\begin{bmatrix}1&-3\\-1&2\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

and both matrices are clearly the same.

More generally, we have by definition of inverse,

(AB)(AB)^{-1} = I

where I is the identity matrix. Multiply on the left by <em>A </em>⁻¹ to get

A^{-1}(AB)(AB)^{-1} = A^{-1}I = A^{-1}

Multiplication of matrices is associative, so we can regroup terms as

(A^{-1}A)B(AB)^{-1} = A^{-1} \\\\ B(AB)^{-1} = A^{-1}

Now multiply again on the left by <em>B</em> ⁻¹ and do the same thing:

B^{-1}\left(B(AB)^{-1}\right) = (B^{-1}B)(AB)^{-1} = B^{-1}A^{-1} \\\\ (AB)^{-1} = B^{-1}A^{-1}

7 0
3 years ago
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