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3 years ago

15

HARD QUADRATICS!!!! A quadratic of the form $-2x^2 + bx + c$ has roots of $x = 3 + \sqrt{5}$ and $x = 3 - \sqrt{5}.$ The graph o

f $y = -2x^2 + bx + c$ is a parabola. Find the vertex of this parabola.

1 answer:

iren2701 [21]3 years ago

5 0

Answer:

hold up this will take a min to work out

Step-by-step explanation:

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Diano4ka-milaya [45]

Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

Math question screen shot down below

Oksanka [162]

Answer:

The correct answer is A, Point P

Step-by-step explanation:

since the additive inverse of 2 is -2 because -2 + 2 = 0 the correct choice must be choice A, Point P.

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3 years ago

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Find the equation of the line passing through points (-1,-6) and (3,5)

Phoenix [80]

Answer:

y=11/4x-13/4

Step-by-step explanation:

Use the slope formula and slope-intercept form y=mx+b to find the equation.

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3 years ago

Solve the quadratic equation 3x²-3x-1=0?<br>(by using the square method)

ohaa [14]

Answer:

x=1 and x=-1/3

Step-by-step explanation:

$3x^2-2x-1=0\\\\3(x^2-\frac{2}{3}x-\frac{1}{3})=0\\ \\3(x^2-\frac{2}{3}x-\frac{1}{3}+\frac{4}{9})=3(\frac{4}{9})\\ \\ 3(x^2-\frac{2}{3}x+\frac{1}{9})=\frac{4}{3}\\ \\ 3(x-\frac{1}{3})^2=\frac{4}{3}\\ \\(x-\frac{1}{3})^2=\frac{4}{9}\\ \\x-\frac{1}{3}=\pm\frac{2}{3}$

$x-\frac{1}{3}=\frac{2}{3}\\ \\x=\frac{3}{3}\\ \\x=1$

$x-\frac{1}{3}=-\frac{2}{3}\\ \\x=-\frac{1}{3}$

Therefore, x=1 and x=-1/3

8 0

3 years ago