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Oksana_A [137]
3 years ago
15

Which property justifies this statement if x=3, then x-3=0

Mathematics
1 answer:
PolarNik [594]3 years ago
5 0

You would subtract 3 from 0 and then the 3's would cancel out and it would be X=3

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Where an, an-1,a2, a1, a0 are constants. We call the term containing the highest power of x the leading term, and we call an the leading coefficient. The degree of the polynomial is the power of x in the leading term. We have already seen degree 0, 1, and 2 polynomials which were the constant, linear, and quadratic functions, respectively. Degree 3, 4, and 5
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What is the equation of the line that is perpendicular to y= -3x + 1 and passes through (2,3)?
Aleonysh [2.5K]

Answer:

\large\boxed{y=\dfrac{1}{3}x+2\dfrac{1}{3}}

Step-by-step explanation:

\text{Let}\ k:y=_1x+b_1\ \text{and}\ l:y=m_2x+b_2.\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\============================\\\\\text{We have}\ y=-3x+1\to m_1=-3.\\\\\text{Therefore}\ m_2=-\dfrac{1}{-3}=\dfrac{1}{3}.\\\\\text{The equation of the searched line:}\ y=\dfrac{1}{3}x+b.\\\\\text{The line passes through }(2,\ 3).

\text{Put the coordinates of the point to the equation.}\ x=2,\ y=3:\\\\3=\dfrac{1}{3}(2)+b\\\\3=\dfrac{2}{3}+b\qquad\text{subtract}\ \dfrac{2}{3}\ \text{from both sides}\\\\b=2\dfrac{1}{3}

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Find the other endpoint of the line segment with the endpoint (-3,3) and midpoint (1,-4)
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The other end point would be (5,-11)
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For question 25 please pick 1,2,3 or 4
Juli2301 [7.4K]

Answer:

4.

$x=\frac{2}{3} \pm  \frac{1}{6}i \sqrt{158}$

Step-by-step explanation:

18x^2-24x+87=0

Using Quadratic Formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-\left(-24\right)\pm \sqrt{\left(-24\right)^2-4\cdot \:18\cdot \:87}}{2\cdot \:18}$

Solving the discriminant:

\Delta = \left(-24\right)^2-4\cdot \:18\cdot \:87\\\Delta = 576-6264\\ \Delta=-5688

$x= \frac{24 \pm \sqrt{-5688} }{36} $

$x= \frac{24 \pm \sqrt{5688i} }{36} $

Once \sqrt{5688} =\sqrt{2^3\cdot \:3^2\cdot \:79}=6\sqrt{158}

$x=\frac{24\pm6\sqrt{158}i}{36}$

Dividing the denominator and numerator by 6

$x=\frac{4\pm \sqrt{158}i}{6}$

Now rewrite it:

$x=\frac{4}{6} \pm  i\frac{\sqrt{158}}{6}$

$x=\frac{2}{3} \pm i \frac{\sqrt{158}}{6}$

or

$x=\frac{2}{3} \pm  \frac{1}{6}i \sqrt{158}$

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3 years ago
Please help me<br> it means a lot
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1. 5
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3. 10
4. 25
5. -11

6.1/4
7. 1
8. -1

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