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3 years ago

15

A rectangle has vertices at these coordinates.(3,6), (3,4), (6,4)What are the coordinates of the fourth vertex of the rectangle?

1 answer:

omeli [17]3 years ago

5 0

Hi, i hope this helps! The coordinates are (6,6). I figured this out by plotting the given points and remembering that a rectangle has parallel sides and right angles :))

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Write the equation of the parabola that has its x-intercepts at (1+<img src="https://tex.z-dn.net/?f=%5Csqrt%7B5%7D" id="TexForm

VladimirAG [237]

Answer:

$y=2x^2-4x-8$

Step-by-step explanation:

<u>Factored form of a parabola</u>

$y=a(x-p)(x-q)$

where:

p and q are the x-intercepts.
a is some constant.

Given x-intercepts:

(1+√5, 0)
(1-√5, 0)

Therefore:

$\implies y=a(x-(1+\sqrt{5}))(x-(1-\sqrt{5}))$

$\implies y=a(x-1-\sqrt{5})(x-1+\sqrt{5})$

To find a, substitute the given point (4, 8) into the equation and solve for a:

$\implies a(4-1-\sqrt{5})(4-1+\sqrt{5})=8$

$\implies a(3-\sqrt{5})(3+\sqrt{5})=8$

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$\implies a=2$

Therefore, the equation of the parabola in factored form is:

$\implies y=2(x-1-\sqrt{5})(x-1+\sqrt{5})$

Expand so that the equation is in standard form:

$\implies y=2(x^2-x+\sqrt{5}x-x+1-\sqrt{5}-\sqrt{5}x+\sqrt{5}-5)$

$\implies y=2(x^2-x-x+\sqrt{5}x-\sqrt{5}x+\sqrt{5}-\sqrt{5}+1-5)$

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2 years ago

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2 years ago

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Answer:

The answer is C

$\frac{4 \times - 4 - 4}{ - 4 - 1}$

$\frac{ - 16 - 4}{ - 5}$

$\frac{ - 20}{ - 5} = 4$

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3 years ago