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Yuki888 [10]
3 years ago
12

Solve the linear equation 1/3x - 5 + 171 = x

Mathematics
1 answer:
sdas [7]3 years ago
5 0

Answer:  x = 249

Step-by-step explanation:

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7 1/2

Step-by-step explanation:

2 1/2 times 3 is 7 1/2

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8a is the common favor of 6a

Step-by-step explanation:

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For the first option, since when x is going down (to the left) the function is going up, it's not approaching 0. For the second option, since when x is going up (to the right) it's going up, it's not approaching negative infinity (negative infinity is all the way down). For the third one, since when x is going down the y values are climbing, we can assume that the function's values go to positive infinity. For the last one, since when x=0 y=0, when x=0 the function does not go to infinity
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What is the value for x in the diagram? Rounded to the nearest whole<br> degree.
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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}.

4 0
3 years ago
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