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3 years ago

Solve for x: -2(x + 3) = -2(x + 1) - 4

Mathematics

2 answers:

Nataly [62]3 years ago

8 0

$-2(x+3)=-2(x+1)-4\\\\-2x-6=-2x-2-4\\\\-2x-6=-2x-6\ \ \ \ |add\ (-2x)\ to\ both\ sides\\\\-6=-6\ \ TRUE\\\\
infinitely\ many\ solutions,\ x\in\mathbb{R}.$

meriva3 years ago

7 0

$-2(x + 3) = -2(x + 1) - 4\\
-2x-6=-2x-2-4\\
-6=-6\\
x\in\mathbb{R}$

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3 years ago

The Country Buffet restaurant has tables that seat 6 people and booths that can seat 4 people. The restaurant has 38 seating uni

Alina [70]

Answer:

There are 20 booths and (38 - 20), or 18, tables

Step-by-step explanation:

Represent the number of tables with t and the number of booths with b.

We need to find the values of t and b.

(6 people/table)(t) + (4 people/booth)b = 188 (units are "people")

t + b = 38 (units are "seating units")

Solving the second equation for t, we get 38 - b = t.

Substitute 38 - b for t in the first equation:

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3 years ago

Factor the polynomial, x2 + 5x + 6

patriot [66]

Answer:

Choice b.

$x^{2} + 5\, x + 6 = (x + 3)\, (x + 2)$ .

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 $(x + 2)$ .
The root $x = -3$ corresponds to the factor $(x - (-3))$ , which simplifies to $(x + 3)$ .

Verify that $(x + 2)\, (x + 3)$ indeed expands to the original polynomial:

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

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