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3 years ago

Solve for X. (1 point) -ax + 3b > 5

Mathematics

1 answer:

Anarel [89]3 years ago

4 0

Answer:

x<(3b-5)/a

Step-by-step explanation:

Subtract 3b from both sides: -ax>5-3b

Divide by -a, when dividing/multiplying a negative you flip the </> sign: x<(3b-5)/a

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Solve the system of linear equations using the Gauss-Jordan elimination method. 2x + 3y 6212 3x + (x, y. z)

Gekata [30.6K]

Answer:

The solution of the system of linear equations is $x=3, y=4, z=1$

Step-by-step explanation:

We have the system of linear equations:

$2x+3y-6z=12\\x-2y+3z=-2\\3x+y=13$

Gauss-Jordan elimination method is the process of performing row operations to transform any matrix into reduced row-echelon form.

The first step is to transform the system of linear equations into the matrix form. A system of linear equations can be represented in matrix form (Ax=b) using a coefficient matrix (A), a variable matrix (x), and a constant matrix(b).

From the system of linear equations that we have, the coefficient matrix is

$\left[\begin{array}{ccc}2&3&-6\\1&-2&3\\3&1&0\end{array}\right]$

the variable matrix is

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

and the constant matrix is

$\left[\begin{array}{c}12&-2&13\end{array}\right]$

We also need the augmented matrix, this matrix is the result of joining the columns of the coefficient matrix and the constant matrix divided by a vertical bar, so

$\left[\begin{array}{ccc|c}2&3&-6&12\\1&-2&3&-2\\3&1&0&13\end{array}\right]$

To transform the augmented matrix to reduced row-echelon form we need to follow these row operations:

multiply the 1st row by 1/2

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\1&-2&3&-2\\3&1&0&13\end{array}\right]$

add -1 times the 1st row to the 2nd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\3&1&0&13\end{array}\right]$

add -3 times the 1st row to the 3rd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\0&-7/2&9&-5\end{array}\right]$

multiply the 2nd row by -2/7

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&-7/2&9&-5\end{array}\right]$

add 7/2 times the 2nd row to the 3rd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&0&3&3\end{array}\right]$

multiply the 3rd row by 1/3

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&0&1&1\end{array}\right]$

add 12/7 times the 3rd row to the 2nd row

$\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&0&4\\0&0&1&1\end{array}\right]$

add 3 times the 3rd row to the 1st row

$\left[\begin{array}{ccc|c}1&3/2&0&9\\0&1&0&4\\0&0&1&1\end{array}\right]$

add -3/2 times the 2nd row to the 1st row

$\left[\begin{array}{ccc|c}1&0&0&3\\0&1&0&4\\0&0&1&1\end{array}\right]$

From the reduced row echelon form we have that

$x=3\\y=4\\z=1$

Since every column in the coefficient part of the matrix has a leading entry that means our system has a unique solution.

7 0

3 years ago

Simplify. 3(x + 4) please

Sveta_85 [38]

Answer:

3 x + 12

Step-by-step explanation:

6 0

3 years ago

The proof that point (1,root 3) lies on the circle that is centered at the origin and contains the point (0,2) is found in the t

AURORKA [14]

<h2>Answer:</h2>

D. Distance formula

<h2>Step-by-step explanation:</h2>

$The \ \mathbf{distance} \ d \ between \ the \ \mathbf{points} \ (x_{1},y_{1}) \ and \ (x_{2},y_{2}) \ in \ the \ plane \ is:\\ \\ d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

As in the statement:

<em>The distance form</em> $(0,0)$ to $(0,2)$ is $\sqrt{(0-0)^2+(2-0)^2}=\sqrt{2^2}=2$ <em>whose justification is the </em><em>Distance Formula. </em>

<em />

Here in this statement the justification is also the Distance Fromula, so we take the distance from $(0,0) \ to \ (1,\sqrt{2})$ whose result is also 2 and is the radius of the circle.

3 0

3 years ago

Read 2 more answers

The coordinates of the vertices of / are G(–2, 3), H(–1, 2), and I(–3, 1). If / is reflected across the x-axis to create /, find

valkas [14]

Answer:

A. (–2, –3)

that is the answer

3 0

3 years ago

One day, eleven babies are born at a hospital. Assuming each baby has an equal chance of being a boy or girl, what is the probab

Andrews [41]

You only need to consider the situations where 10 or 11 of the babies are girls, then subtract those probabilities from 1. This will give probability that any other number up to 9 of the babies are girls.

Use binomial theorem.
$P(x=k) = (nCk) p^k (1-p)^{n-k}$

n = 11
k = 10,11
p = 1/2

$P(x=10) = 11 (\frac{1}{2})^{11} = \frac{11}{2048} \\ \\ P(x=11) = 1(\frac{1}{2})^{11} = \frac{1}{2048} \\ \\ P(x \leq 9) = 1 - \frac{11}{2048} - \frac{1}{2048} \\ \\ P(x \leq 9)=\frac{2036}{2048} = \frac{509}{512}$

4 0

3 years ago