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2 years ago

HELP HELP!!

Mathematics

1 answer:

Goryan [66]2 years ago

7 0

Answer:

Step-by-step explanation:

$p_1:~~y = x^2+2\\p_2:~~y = 3x^2+2\\ \\ V{p_1} = \Big(-\dfrac{b}{2a}, -\dfrac{\Delta}{4a}\Big) = \Big(-\dfrac{0}{2}, -\dfrac{0^2-4\cdot 2}{4}\Big) = \Big(0,2\Big) \\ \\ Vp_2 = \Big(x_V, -\dfrac{\Delta}{4a}\Big) = \Big(0, -\dfrac{0^2-4\cdot 3 \cdot 2}{4\cdot 3}\Big) = \Big(0,2\Big) \\ \\ \\ \text{The distance is }0,~~\text{Because the vertices are equal.}$

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Step-by-step explanation:

We have that:

$\frac{x^{a}}{x^{b}} = x^{a-b}$

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Then

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Y = x² - 4x + 4
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Find intersection of L and C:
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x² - 6x + 8 = 0
<span> (x - 2)(x - 4) = 0
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When x = 2 , y = 2(2) - 4 = 0
When x = 4, y = 2(4) - 4 = 4

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</span> $
\text {Length of AB}= \sqrt{(4-0)^2 + (4-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 4.47 \text{ units}$
<span>
Answer: 4.47 units

</span>

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