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Liono4ka [1.6K]
3 years ago
13

Pls help ill give u brainlest

Mathematics
1 answer:
wlad13 [49]3 years ago
6 0

Answer:

square

Step-by-step explanation:

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Solve:<br> 3x−5 = 9. <br> A. x = −3 <br> B.x = 2 <br> C.x = 7 <br> D.x = 8
Alina [70]

Answer:

Step-by-step explanation:

3x-5=9

3x-5+5=9+5

3x=14

\frac{3x}{3}=\frac{14}{3}

x=\frac{14}{3}

x ≈ 4.66667

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3 years ago
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sergey [27]
Here you go!! Hope this helps

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Solve the following systems of equations using the matrix method: a. 3x1 + 2x2 + 4x3 = 5 2x1 + 5x2 + 3x3 = 17 7x1 + 2x2 + 2x3 =
lara [203]

Answer:

a. The solutions are

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}

b. The solutions are

\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}

c. The solutions are

\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]=\begin{pmatrix}\frac{22}{9}\\ \frac{164}{9}\\ \frac{139}{9}\\ -\frac{37}{3}\end{pmatrix}

Step-by-step explanation:

Solving a system of linear equations using matrix method, we may define a system of equations with the same number of equations as variables as:

A\cdot X=B

where X is the matrix representing the variables of the system,  B is the matrix representing the constants, and A is the coefficient matrix.

Then the solution is this:

X=A^{-1}B

a. Given the system:

3x_1 + 2x_2 + 4x_3 = 5 \\2x_1 + 5x_2 + 3x_3 = 17 \\7x_1 + 2x_2 + 2x_3 = 11

The coefficient matrix is:

A=\left[\begin{array}{ccc}3&2&4\\2&5&3\\7&2&2\end{array}\right]

The variable matrix is:

X=\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]

The constant matrix is:

B=\left[\begin{array}{c}5&17&11\\\end{array}\right]

First, we need to find the inverse of the A matrix. To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

\left[ \begin{array}{ccc|ccc}3&2&4&1&0&0 \\\\ 2&5&3&0&1&0 \\\\ 7&2&2&0&0&1\end{array}\right]

This matrix can be transformed by a sequence of elementary row operations to the matrix

\left[ \begin{array}{ccc|ccc}1&0&0&- \frac{2}{39}&- \frac{2}{39}&\frac{7}{39} \\\\ 0&1&0&- \frac{17}{78}&\frac{11}{39}&\frac{1}{78} \\\\ 0&0&1&\frac{31}{78}&- \frac{4}{39}&- \frac{11}{78}\end{array}\right]

And the inverse of the A matrix is

A^{-1}=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right]

Next, multiply A^ {-1} by B

X=A^{-1}\cdot B

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right] \cdot \left[\begin{array}{c}5&17&11\end{array}\right]

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}-\frac{2}{39}&-\frac{2}{39}&\frac{7}{39}\\ -\frac{17}{78}&\frac{11}{39}&\frac{1}{78}\\ \frac{31}{78}&-\frac{4}{39}&-\frac{11}{78}\end{pmatrix}\begin{pmatrix}5\\ 17\\ 11\end{pmatrix}=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}

The solutions are

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}

b. To solve this system of equations

x -y - z = 0 \\30x + 40y = 12 \\30x + 50z = 12

The coefficient matrix is:

A=\left[\begin{array}{ccc}1&-1&-1\\30&40&0\\30&0&50\end{array}\right]

The variable matrix is:

X=\left[\begin{array}{c}x&y&z\\\end{array}\right]

The constant matrix is:

B=\left[\begin{array}{c}0&12&12\\\end{array}\right]

The inverse of the A matrix is

A^{-1}=\left[ \begin{array}{ccc} \frac{20}{47} & \frac{1}{94} & \frac{2}{235} \\\\ - \frac{15}{47} & \frac{4}{235} & - \frac{3}{470} \\\\ - \frac{12}{47} & - \frac{3}{470} & \frac{7}{470} \end{array} \right]

The solutions are

\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}

c. To solve this system of equations

4x_1 + 2x_2 + x_3 + 5x_4 = 0 \\3x_1 + x_2 + 4x_3 + 7x_4 = 1\\ 2x_1 + 3x_2 + x_3 + 6x_4 = 1 \\3x_1 + x_2 + x_3 + 3x_4 = 4\\

The coefficient matrix is:

A=\left[\begin{array}{cccc}4&2&1&5\\3&1&4&7\\2&3&1&6\\3&1&1&3\end{array}\right]

The variable matrix is:

X=\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]

The constant matrix is:

B=\left[\begin{array}{c}0&1&1&4\\\end{array}\right]

The inverse of the A matrix is

A^{-1}=\left[ \begin{array}{cccc} - \frac{1}{9} & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{3} \\\\ - \frac{32}{9} & - \frac{5}{9} & \frac{13}{9} & \frac{13}{3} \\\\ - \frac{28}{9} & - \frac{1}{9} & \frac{8}{9} & \frac{11}{3} \\\\ \frac{7}{3} & \frac{1}{3} & - \frac{2}{3} & -3 \end{array} \right]

The solutions are

\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]=\begin{pmatrix}\frac{22}{9}\\ \frac{164}{9}\\ \frac{139}{9}\\ -\frac{37}{3}\end{pmatrix}

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3 years ago
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carla's vacation will cost at least $718. she saves $85 each week. she already has $463. how many weeks,w,does she have to save
aleksklad [387]
She already has 463 that's a set amount she started at an her start amount won't change an she's wanting 718 an she is savin 85 a week so every week she gets more money so that number changes so ur equation is 85W+463=718 then u do a two step equation
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Geometry help please
Lady_Fox [76]
I would say that the answer is A. Parallel because, the points are each the same distance away (horizontally) going by the little faded dots in the background
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