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3 years ago

Math please help im failing this class its algebra

Mathematics

2 answers:

12345 [234]3 years ago

5 0

Answer:

y=5x+2

Step-by-step explanation:

-5x-4y=-8

-5x+y=2

y=2+5x

y=5x+2

astra-53 [7]3 years ago

4 0

Answer:

The answer would be zero 0

Step-by-step explanation:

1. move the constant to the left-hand side and changes it's sign

-5x-4y=-8 → -5x-4y+8=0

2. changes the signs of an both sides of the equation

-5x-4y+8=0 → 5x+4y-8 = 0

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Please simplify, 50 points.

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Answer:

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Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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7 0

3 years ago

Find a fundamental matrix solution for the system x 0 1 7x1 + 4x2 + 12x3, x 0 2 x1 + 2x2 + x3, x 0 3 −3x1 − 2x2 − 5x3. Then find

liq [111]

Here is the correct format for the question.

$x'_1 = 7x_1 +4x_2+ 12x_3 , \ \ x'_2 = x_1 + 2x_2 + x_3 , \ \ x'_3 = -3x_1 -2x_2 -5x_3$ . Then find the solution that satisfies $x \limits ^{\to} = \left[\begin{array}{c}0\\1\\-2\end{array}\right]$

Answer:

$\mathbf{c_1 = -3, c_2 = 2, c_3 = 2}$

Step-by-step explanation:

From the figures given above:

the matrix can be computed as,

$\left[\begin{array}{ccc}7&4&12\\1&2&1\\-3&-2&-5\end{array}\right] x \limits ^{\to} = \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] = x \limits ^{\to} = \left[\begin{array}{c}x_1'\\x_2'\\x_3'\end{array}\right]$

The first thing we need to carry out is to determine the eigenvalues of A,

where:

$A = \left[\begin{array}{ccc}7&4&12\\1&2&1\\-3&-2&-5\end{array}\right]$

$|A-rI|=0$

$\begin {vmatrix} 7-r&4&12\\1&2-r&1\\-3&-2&-5-r \end {vmatrix}=0$

the eigenvalues are r = 0, 1, 3

However, the eigenvector correlated to each eigenvalue can be calculated as follows.

suppose r = 0

(A - rI) x = 0

$\left[\begin{array}{ccc}7&4&12\\1&2&1\\-3&-2&-5\end{array}\right] \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] = \left[\begin{array}{c}0\\0\\0\end{array}\right]$

now the eigenvector is $\left[\begin{array}{c}-4\\1\\2\end{array}\right]$

However, for eigenvalue = 1, we have : $\left[\begin{array}{c}-4\\1\\2\end{array}\right]$

for eigenvalue = 3, we have: $\left[\begin{array}{c}-2\\-1\\1\end{array}\right]$

The solution now can be computed as :

$x(t)= c_1 \left[\begin{array}{c}-4\\1\\2\end{array}\right] + c_2e^t \left[\begin{array}{c}-4\\3\\1\end{array}\right]+ c_3e^{3t} \left[\begin{array}{c}-2\\-1\\1\end{array}\right]$

Similarly, the fundamental matrix solution is:

$\left[\begin{array}{ccc}-4&-4e^t&-2e^{3t}\\1&3e^t&-e^{3t}\\2&e^t&e^{3t}\end{array}\right]$

$-4c_1 -4c_2-2c_3 =0 \\ \\ c_1 + 3c_2 -c_3 = 1\\ \\ 2c_1+c_2 +c_3 = -2$

Solving the above equation, we get:

$\mathbf{c_1 = -3, c_2 = 2, c_3 = 2}$

5 0

3 years ago

2. Trapezoid WXYZ is located at W(2.1) X[3.

Nataliya [291]

Point Y is (6, -1)

Subtract 6 from the y coordinate in point y since it is translated down. keep the x coordinate since it is only being translated up and down, or by the y axis

5 0

4 years ago

The difference of 3 times a number and 15 is no less than the square of the number.

Citrus2011 [14]

Answer:

Answer: 3x-15 greater than/= x^2

Step-by-step explanation:

6 0

4 years ago