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noname [10]
2 years ago
15

I need to find x for these

Mathematics
1 answer:
Ludmilka [50]2 years ago
3 0

Answer:

1. x = 2

2. x = -2

3 x = 1

Step-by-step explanation:

Make x the subjects and work from there

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Step-by-step explanation:

827/200

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-4x - 6y = 6<br> 4x + 6y = -4
Sloan [31]

Step-by-step explanation:

-4x - 6y = 6

4x + 6y = -4

To solve a system of equations, we can add the two equations and solve for one of the remaining variables -- let's try to eliminate the x variable when we add the two equations together.

Right now, there's a -4x term in the first equation, and a 4x term in the second equation, so if we add those together, we'll be able to eliminate the x variable altogether and solve for y.

However, when we also have a -6y term in the first equation and 6y term in the second equation, so adding these together will also eliminate the y term, leaving a 0 on the left-hand side of the equation.

If we add the two numbers on the right side of the equation, we get -2, which does not equal 0, meaning there are no solutions to this system of equations.

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2 years ago
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EleoNora [17]

Answer:

I'm lost I messed up how do I delete my answer...

6 0
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Can someone help me in this plssss I really need it
kompoz [17]

Answer:

\boxed{-3xy^{2}\sqrt [3] {2x^{2}}}

Step-by-step explanation:

Your expression is

\sqrt [3] {-54x^{5}y^{6}}

Here's how I would simplify it.

\begin{array}{rcll}\sqrt [3] {-54x^{5}y^{6}} & = & \sqrt [3] {(-1)^{3}\times 2 \times 27 \times x^{2} \times x^{3} \times y^{6}} & \text{Factored the cubes}\\& = & \sqrt [3] {(-1)^{3} \times 3^{3}\times x^{3} \times y^{6}\times 2 \times x^{2}} & \text{Grouped the cubes}\\\end{array}

\begin{array}{rcll}& = & \sqrt [3] {(-1)^{3} \times {3^{3}\times x^{3} \times y^{6}}} \times\sqrt [3] { 2 \times x^{2}} & \text{Separated the cubes}\\&=& \mathbf{-3xy^{2}\sqrt [3] {2x^{2}}} & \text{Took cube roots}\\\end{array}

\text{The simplified expression is $\boxed{\mathbf{-3xy^{2}\sqrt [3] {2x^{2}}}}$}

6 0
3 years ago
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