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prisoha [69]
3 years ago
11

Please help :) If you know anything about combining equations can you help me :)

Mathematics
2 answers:
natka813 [3]3 years ago
7 0

Answer:

-8y-5x=-5

Step-by-step explanation:

Tocombine equations add the two sides together.

(4x-4y)+(-9x-4y)=-2+(-3)

4x-4y-9x-4y= -5

-8y-5x=-5

julsineya [31]3 years ago
5 0

Answer:

-5x - 8y = -5

Step-by-step explanation:

Whenever we two equations, we are essentially just adding the left sides up and then the right sides up. Here, let's write these into one whole big equation:

(4x - 4y) + (-9x - 4y) = -2 + (-3)

Taking the parentheses out, we have:

4x - 4y - 9x - 4y = -2 - 3

Combine like terms:

4x - 9x - 4y - 4y = -5

-5x - 8y = -5

Hope this helps!

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4 years ago
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Answer:

Choice b.

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Step-by-step explanation:

The highest power of the variable x in this polynomial is 2. In other words, this polynomial is quadratic.

It is thus possible to apply the quadratic formula to find the "roots" of this polynomial. (A root of a polynomial is a value of the variable that would set the polynomial to 0.)

After finding these roots, it would be possible to factorize this polynomial using the Factor Theorem.

Apply the quadratic formula to find the two roots that would set this quadratic polynomial to 0. The discriminant of this polynomial is (5^{2} - 4 \times 1 \times 6) = 1.

\begin{aligned}x_{1} &= \frac{-5 + \sqrt{1}}{2\times 1} \\ &= \frac{-5 + 1}{2} \\ &= -2\end{aligned}.

Similarly:

\begin{aligned}x_{2} &= \frac{-5 - \sqrt{1}}{2\times 1} \\ &= \frac{-5 - 1}{2} \\ &= -3\end{aligned}.

By the Factor Theorem, if x = x_{0} is a root of a polynomial, then (x - x_0) would be a factor of that polynomial. Note the minus sign between x and x_{0}.

  • The root x = -2 corresponds to the factor (x - (-2)), which simplifies to (x + 2).
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Verify that (x + 2)\, (x + 3) indeed expands to the original polynomial:

\begin{aligned}& (x + 2)\, (x + 3) \\ =\; & x^{2} + 2\, x + 3\, x + 6 \\ =\; & x^{2} + 5\, x + 6\end{aligned}.

4 0
3 years ago
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Step-by-step explanation:

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