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DiKsa [7]
2 years ago
9

What is the solution to the system of equations below? 3x+4y-2z=-1 -x+6y+2z=7 4x-2y+3z=27

Mathematics
1 answer:
matrenka [14]2 years ago
6 0

Answer:

(3x+4y-2z=-1)= x=-1/3-4/3y+2/3z

(-x+6y=2z)= x=-7+6y=2z

(4x-2y+3z=27)= x=27/4+1/2-3/4z

Step-by-step explanation:

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Answer: OPTION D.

Step-by-step explanation:

<h3> The complete exercise is: "Which arrangement shows 3\frac{1}{8}, 3.7, 3\frac{3}{4}, and 3.89 in order from least to greatest?"</h3><h3> </h3>

Convert from mixed numbers to decimal numbers.

The steps to do this, are the following:

1. You must divide the numerator of the fraction by the denominator.

2. Then you must add the quotient obtained to the whole number part.

Applying this procedure, you get that:

3\frac{1}{8}=3+0.125=3.125

3\frac{3}{4}=3+0.75=3.75

Now that you have all the numbers in decimal form, you can order from least to greatest:

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Therefore, you can conclude that the correct arrangement is:

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3 years ago
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GarryVolchara [31]

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Step-by-step explanation:

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(8y²)(-3x²y²)(2/3xy⁴)<br><br> HELP PLEASEEEEEEEE
IRINA_888 [86]

Step-by-step explanation:

1 Remove parentheses.

8{y}^{2}\times -3{x}^{2}{y}^{2}\times \frac{2}{3}x{y}^{4}

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2

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y

2

×

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2

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2 Use this rule: \frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}

b

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bd

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\frac{8{y}^{2}\times -3{x}^{2}{y}^{2}\times 2x{y}^{4}}{3}

3

8y

2

×−3x

2

y

2

×2xy

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3 Take out the constants.

\frac{(8\times -3\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}

3

(8×−3×2)y

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y

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y

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x

2

x

4 Simplify 8\times -38×−3 to -24−24.

\frac{(-24\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}

3

(−24×2)y

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y

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x

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\frac{-48{y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}

3

−48y

2

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2

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x

2

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6 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x

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x

b

=x

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\frac{-48{y}^{2+2+4}{x}^{2+1}}{3}

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\frac{-48{y}^{4+4}{x}^{2+1}}{3}

3

−48y

4+4

x

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8 Simplify 4+44+4 to 88.

\frac{-48{y}^{8}{x}^{2+1}}{3}

3

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8

x

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9 Simplify 2+12+1 to 33.

\frac{-48{y}^{8}{x}^{3}}{3}

3

−48y

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3

10 Move the negative sign to the left.

-\frac{48{y}^{8}{x}^{3}}{3}

−

3

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8

x

3

11 Simplify \frac{48{y}^{8}{x}^{3}}{3}

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to 16{y}^{8}{x}^{3}16y

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-16{y}^{8}{x}^{3}

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