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3 years ago

5

Write an equation of a line in point-slope and slope intercept form that passes through the point (1, -8) and is perpendicular t

o the line: 2x-4y = -11

1 answer:

prisoha [69]3 years ago

3 0

Answer:

11y-2x= -90

Step-by-step explanation:

for the point (1,-8)

x=1 and y= -8

for the equation,

2x -4y= -11

x= -1/m

m= -1/x

x = -1/ (-11/2 )

m= 2/11

so for the equation,

y-y1= m(x- x1)

y- -8=2/11(x-1)

11(y+8)= 2(x-1)

11y +88= 2x-2

11y-2x= -90

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<u>ANSWER: </u>

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

<u>SOLUTION:</u>

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

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$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$

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