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MArishka [77]
3 years ago
6

Write the augmented matrix for the following system of equations.

Mathematics
1 answer:
Llana [10]3 years ago
6 0
62

27

68 it’s very simple do the math
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Which ordered pairs are solutions to the inequality −2x+y≥−4 ?
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3 years ago
Find the solution of the system of equations.<br> 15x – 4y = -50<br> 3x – 2y = –16
rusak2 [61]

Answer:

x=-2, y=5. (-2, 5).

Step-by-step explanation:

15x-4y=-50

3x-2y=-16

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

15x-4y=-50

-5(3x-2y)=-5(-16)

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

15x-4y=-50

-15x+10y=80

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

6y=30

y=30/6

y=5

3x-2(5)=-16

3x-10=-16

3x=-16+10

3x=-6

x=-6/3

x=-2

7 0
3 years ago
Solve the system by using a matrix equation.<br> --4x - 5y = -5<br> -6x - 8y = -2
evablogger [386]

Answer:

Solution : (15, - 11)

Step-by-step explanation:

We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )

\begin{bmatrix}-4&-5&|&-5\\ -6&-8&|&-2\end{bmatrix}

Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )

Row Echelon Form :

\begin{pmatrix}1\:&\:\cdots \:&\:b\:\\ 0\:&\ddots \:&\:\vdots \\ 0\:&\:0\:&\:1\end{pmatrix}

Step # 1 : Swap the first and second matrix rows,

\begin{pmatrix}-6&-8&-2\\ -4&-5&-5\end{pmatrix}

Step # 2 : Cancel leading coefficient in row 2 through R_2\:\leftarrow \:R_2-\frac{2}{3}\cdot \:R_1,

\begin{pmatrix}-6&-8&-2\\ 0&\frac{1}{3}&-\frac{11}{3}\end{pmatrix}

Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.

\begin{bmatrix}1&0&|&15\\ 0&1&|&-11\end{bmatrix}

As you can see our solution is x = 15, y = - 11 or (15, - 11).

4 0
3 years ago
If 3 servings is 5 cups, how many servings is 15 cups
blsea [12.9K]
9 servings, 5×3=15 but for each 5 equals 3 so,3 +3+3 =9 because 5 is a serving
4 0
3 years ago
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