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FromTheMoon [43]
4 years ago
5

-2(x - 4) = -8 solve

Mathematics
2 answers:
Vladimir [108]4 years ago
5 0

Answer:

x=4/-4 - 4/-4

Step-by-step explanation:

lilavasa [31]4 years ago
3 0

X = 8

hope it helps : )

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Pleaseeeee help :'))))
jeka57 [31]

Answer:

The answer is

x= -35

Step-by-step explanation:

x/5= -7

x= -35. :-)

7 0
3 years ago
Read 2 more answers
What is the square root of 43
Tanzania [10]
Square root of 43 is 6.5574 etc
8 0
4 years ago
Read 2 more answers
Is anyone able to help me with this please?
Lapatulllka [165]

9514 1404 393

Answer:

  • range: -2 ≤ y
  • domain: All reals

Step-by-step explanation:

The range of the function is the vertical extent. Here, the values of y can be anything that is -2 or more:

  range: -2 ≤ y

The domain of a the function is the horizontal extent. The domain of <em>any</em> polynomial function is ...

  domain: All reals

8 0
3 years ago
How many solutions does y=5x+12 and y=5x+18
Mrrafil [7]

Answer:

there is no solution for these linear equations in two variables.

since a1 (y) = a2 (y)

         b1 (5x) = b2 (5x)

  but, c1 (12) not equal to c2 (18)

8 0
3 years ago
Read 2 more answers
Can you tell me how to calculate R for below Matrix ?
scoray [572]

Answer:

The rank of the matrix is 3.

Step-by-step explanation:

Consider the prided information.

x=\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}=\begin{bmatrix}9&2&6&5&8\\ 12&8&6&4&10\\ 3&4&0&2&1\end{bmatrix}

Reduce the matrix in row echelon form as shown:

R_1\:\leftrightarrow \:R_2\begin{bmatrix}12&8&6&4&10\\ 9&2&6&5&8\\ 3&4&0&2&1\end{bmatrix}\\

R_2\:\leftarrow \:R_2-\frac{3}{4}\cdot \:R_1\ and\ R_3\:\leftarrow \:R_3-\frac{1}{4}\cdot \:R_1\\\\\begin{bmatrix}12&8&6&4&10\\ 0&-4&\frac{3}{2}&2&\frac{1}{2}\\ 0&2&-\frac{3}{2}&1&-\frac{3}{2}\end{bmatrix}

R_3\:\leftarrow \:R_3+\frac{1}{2}\cdot \:R_2\\\begin{bmatrix}12&8&6&4&10\\ 0&-4&\frac{3}{2}&2&\frac{1}{2}\\ 0&0&-\frac{3}{4}&2&-\frac{5}{4}\end{bmatrix}

R_2\:\leftarrow \:R_2-\frac{3}{2}\cdot \:R_3\ and\ R_1\:\leftarrow \:R_1-6\cdot \:R_3\\\begin{bmatrix}12&8&0&20&0\\ 0&-4&0&6&-2\\ 0&0&1&-\frac{8}{3}&\frac{5}{3}\end{bmatrix}\\R_2\:\leftarrow \:-\frac{1}{4}\cdot \:R_2\ , \ R_1\:\leftarrow \:R_1-8\cdot \:R_2\ and\ \:R_1\:\leftarrow \frac{1}{12}\cdot \:R_1\\\begin{bmatrix}1&0&0&\frac{8}{3}&-\frac{1}{3}\\ 0&1&0&-\frac{3}{2}&\frac{1}{2}\\ 0&0&1&-\frac{8}{3}&\frac{5}{3}\end{bmatrix}

The rank of a matrix is the number of non zeros rows.

Thus, the rank of the matrix is 3.

6 0
3 years ago
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