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Ipatiy [6.2K]
3 years ago
13

Hwlp me with this one plzzzzzzzzzzzz

Mathematics
2 answers:
Genrish500 [490]3 years ago
8 0
Divide both sides by 17
x > -1
blondinia [14]3 years ago
5 0

Answer:

x=1

Step-by-step explanation:

I'm not positive, it looks like a trick question but id say x=1 bc 17 × 1 is 17 and 17 is greater than -17 but then again any number greater than 1 such as 2,3,4, or 5 could work

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PLEASE HELP!!!!
Lunna [17]

NOTES:

Given a quadratic function in standard format (y = ax² + bx + c), the direction of the parabola is as follows:

  • if "a" is positive, then opens UP
  • if "a" is negative, then opens DOWN

Given a quadratic function in standard format (y = ax² + bx + c), the vertex can be found as follows:

  • the Axis Of Symmetry (x-value) is: x = \frac{-b}{2a}  
  • y-value is found by plugging in the AOS for "x" in the equation

****************************************************************************************

1) y = x² + 11x + 24

  • a = +1 so the parabola opens UP
  • x = \frac{-b}{2a} = \frac{-11}{2(1)}  = -\frac{11}{2}
  • y = (-\frac{11}{2})^{2} + 11(-\frac{11}{2} ) + 24 = -\frac{25}{4}
  • vertex (-\frac{11}{2}, -\frac{25}{4}) is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-3, 0) and (-8, 0), and y-intercept (0, 24)

******************************************************************************************

2) y = -x² - 6x - 8

  • a = -1 so the parabola opens DOWN
  • x = \frac{-b}{2a} = \frac{-(-6)}{2(-1)}  = \frac{6}{-2} = -3
  • y = -(-3)² - 6(-3) - 8 = -9 + 18 - 8 = 1
  • vertex (-3, 1) is in Quadrant 2 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-4, 0), and y-intercept (0, -8)

******************************************************************************************

3) y = x² - 2x + 3

  • a = 1 so the parabola opens UP
  • x = \frac{-b}{2a} = \frac{-(-2)}{2(1)}  = \frac{2}{2} = 1
  • y = (1)² - 2(1) + 3 = 1 - 2 + 3 = 2
  • vertex (1, 2) is in Quadrant 1 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 3), and its mirror image (2, 3). <em>There are no x-intercepts</em>

******************************************************************************************

4) y = x² + 4x + 4

  • a = 1 so the parabola opens UP
  • x = \frac{-b}{2a} = \frac{-4}{2(1)}  = -2
  • y = (-2)² + 4(-2) + 4 = 4 - 8 + 4 = 0
  • vertex (-2, 0) is in Quadrant 2 and is on the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 4), and its mirror image (-4, 4). <em>The x-intercept is the vertex.</em>

Compared to the other four graphs, this is most likely the equation for the rain gauge!

******************************************************************************************

5) y = 3x² + 21x + 30

  • a = +3 so the parabola opens UP
  • x = \frac{-b}{2a} = \frac{-21}{2(3)}  = -\frac{7}{2}
  • y = 3(-\frac{7}{2})^{2} + 21(-\frac{7}{2} ) + 30 = -\frac{27}{4}
  • vertex (-\frac{7}{2}, -\frac{27}{4}) is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-5, 0), and y-intercept (0, 30)

*******************************************************************************************

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Rationalize the denominator of 4/14 root 2?
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First of all, you can simplify the 4 at the numerator and the 14 at the denominator (they're both multiple of 2):

\dfrac{4}{14\sqrt{2}} = \dfrac{2\cdot 2}{2\cdot 7\cdot\sqrt{2}} = \dfrac{2}{7\sqrt{2}}

Now, rationalize a denominator means that you have to get rid of the square root, in order to have an integer denominator.

To do so, remember that you can always multiply any number by 1 without changing its value, and you can always think of 1 as a fraction where numerator and denominator are equal:

\dfrac{2}{7\sqrt{2}} = \dfrac{2}{7\sqrt{2}}\cdot 1 = \dfrac{2}{7\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{2\sqrt{2}}{7\cdot 2} = \dfrac{\sqrt{2}}{7}

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