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Maurinko [17]
2 years ago
15

HELP PLZ DUE RN BRAINIEST TO WHOEVER RIGHT

Mathematics
2 answers:
iogann1982 [59]2 years ago
6 0

Answer:

Answer:

x = \dfrac{2}{5} + \dfrac{\sqrt{14}}{5}x=

5

2

+

5

14

or x = \dfrac{2}{5} - \dfrac{\sqrt{14}}{5}x=

5

2

−

5

14

Step-by-step explanation:

5x^2 - 2 = 4x5x

2

−2=4x

5x^2 - 4x - 2 = 05x

2

−4x−2=0

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

2a

−b±

b

2

−4ac

x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(5)(-2)}}{2(5)}x=

2(5)

−(−4)±

(−4)

2

−4(5)(−2)

x = \dfrac{4 \pm \sqrt{16 + 40}}{10}x=

10

4±

16+40

x = \dfrac{4 \pm 2\sqrt{14}}{10}x=

10

4±2

14

x = \dfrac{2 \pm \sqrt{14}}{5}x=

5

2±

14

x = \dfrac{2}{5} + \dfrac{\sqrt{14}}{5}x=

5

2

+

5

14

or x = \dfrac{2}{5} - \dfrac{\sqrt{14}}{5}x=

5

2

−

5

14

Y_Kistochka [10]2 years ago
4 0

Answer:

x = \dfrac{2}{5} + \dfrac{\sqrt{14}}{5}   or   x = \dfrac{2}{5} - \dfrac{\sqrt{14}}{5}

Step-by-step explanation:

5x^2 - 2 = 4x

5x^2 - 4x - 2 = 0

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

x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(5)(-2)}}{2(5)}

x = \dfrac{4 \pm \sqrt{16 + 40}}{10}

x = \dfrac{4 \pm 2\sqrt{14}}{10}

x = \dfrac{2 \pm \sqrt{14}}{5}

x = \dfrac{2}{5} + \dfrac{\sqrt{14}}{5}   or   x = \dfrac{2}{5} - \dfrac{\sqrt{14}}{5}

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<h3>Answer:  Choice D.   (3j, 3k)  and (3/j, 3/k)</h3>

The slash indicates a fraction.

=============================================

Proof:

We'll need to consider 4 different cases.

-----------------------

Case (1): j > 0 and k > 0

If j > 0, then 3j > 0 and 3/j > 0

If k > 0, then 3k > 0 and 3/k > 0

The two points (3j, 3k) and (3/j, 3/k) are both in quadrant 1.

-----------------------

Case (2): j > 0 and k < 0

If j > 0, then 3j > 0 and 3/j > 0

If k < 0, then 3k < 0 and 3/k < 0

Points (3j, 3k) and (3/j, 3/k) are both in quadrant 4.

------------------------

Case (3): j < 0 and k > 0

If j < 0, then 3j < 0 and 3/j < 0

If k > 0, then 3k > 0 and 3/k > 0

Points (3j, 3k) and (3/j, 3/k) are in quadrant 3.

------------------------

Case (4): j < 0 and k < 0

If j < 0, then 3j < 0 and 3/j < 0

If k < 0, then 3k < 0 and 3/k < 0

Points (3j, 3k) and (3/j, 3/k) are in quadrant 4.

------------------------

For nonzero integers j and k, we've shown that Points (3j, 3k) and (3/j, 3/k) are in the same quadrant. This concludes the proof.

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2 years ago
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mihalych1998 [28]
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6 0
3 years ago
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tatuchka [14]

Answer:

The slope of the line is 3 ---> m = 3

Step-by-step explanation:

To find the slope of the line, you can choose any two given points from the table. Let's do (1, 3) and (2, 6)

Slope formula:  

m = (y2 - y1) / (x2 - x1)

m = (6 - 3) / (2 - 1)

m = 3 / 1

m = 3

3 0
3 years ago
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