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Kipish [7]
3 years ago
15

Which matrix best represents the system of equations below?

Mathematics
1 answer:
Effectus [21]3 years ago
6 0

\left\{\begin{array}{ccc}-10x-9y+13z=12\\8x-7y-5z=15\\5x+4y-6z=-12\end{array}\right\\\\\begin{array}{ccc}\ \ \ \ x&\ \ \ y&\ \ z\end{array}\\\left[\begin{array}{ccc|c}-10&-9&13&12\\8&-7&-5&15\\5&4&-6&-12\end{array}\right]

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~ H E L P ! ! !~
alex41 [277]

Answer:

5 pounds Banana , and five pound apples

the price of banana=0.8*5=4 dollars

the price of apples = 1.4*5=7 dollars

4+7=11 dollars

no Janice's mother did not give her enough money, she still needs one dollar.

( short one dollar)

8 0
3 years ago
Look at number 17. What did I do wrong?
Gnesinka [82]
When you are checking your answers, your formula should read out as 9(2.222) - 8 = 12, which is simply filing in the x value that you calculated (it comes out to 11.998 which is correct). You wrote it out as 9(2.22) - 8 = -6, which is why your checking process was incorrect 
7 0
3 years ago
Read 2 more answers
X intercept for 2x +5y = -6
Vlad1618 [11]
2x + 5y = -6
-5y -5y
___________
2x = -11y
__ ___
2 2
x = 5.5
8 0
3 years ago
Read 2 more answers
A ball will be drawn from the bag containing 20 balls numbered one through 20 if each ball is equally likely to be drawn what is
son4ous [18]

Answer:

3/5

Step-by-step explanation:

There are 20 balls, numbered 1 through 20.

Of these, 10 are even, 4 are less than five, and 2 are both even and less than 5.

So the probability is:

P(even or <5) = P(even) + P(<5) − P(even and <5)

P(even or <5) = 10/20 + 4/20 − 2/20

P(even or <5) = 12/20

P(even or <5) = 3/5

5 0
3 years ago
Use a matrix equation to solve the system of linear equations. left brace Start 2 By 1 Matrix 1st Row 1st Column 2nd Row 1st Col
natali 33 [55]

Answer:

\left[\begin{array}{ccc}3&-1&\\-1&1/2\\\end{array}\right]

Step-by-step explanation:

The matrix system for the linear equations: x + 2y = 8, 2x + 6y = 9

\left[\begin{array}{ccc}1&2&\\2&6\\\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}8\\9\end{array}\right]

To get the coefficient of x and y, the inverse of the first matrix (let the first matrix be A) must be known.

A^{-1} = (1 / determinant of A) x Adjoint of A

the determinant of A = (1 x 6) - (2 x 2) = 6 - 4 = 2

Adjoint of A = \left[\begin{array}{ccc}6&-2&\\-2&1\\\end{array}\right]

A^{-1}= \frac{1}{2} \left[\begin{array}{ccc}6&-2\\-2&1\end{array}\right] = \left[\begin{array}{ccc}3&-1&\\-1&1/2\\\end{array}\right]

4 0
2 years ago
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