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Dafna1 [17]
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
Mathematics
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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Lapatulllka [165]

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3 years ago
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Len [333]

Answer:

36.45

Step-by-step explanation:

tan 32 = a/35

x = 23.1

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x^2 = a^2+b^2

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8 0
2 years ago
Solve the System of Equations below using Substitution or Elimination
11Alexandr11 [23.1K]

Answer:

(0,0) or Infinitely many solutions.

Step-by-step explanation:

-−4x−4y=0

−4x−4y+4y=0+4y(Add 4y to both sides)

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(Divide both sides by -4)

x=−y

4x+4y=0

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7 0
2 years ago
Determine the quotient of 3 over 7 divided by 2 over 3 . 6 over 21 1 over 2 9 over 14 1 and 5 over 9
Naddika [18.5K]

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8 1
3 years ago
Identify the x-intercepts of the function below f(x)=x^2+12x+24
damaskus [11]

<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.

X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Here, for (1) a = 1, b= 12 and c = 24

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}

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

8 0
2 years ago
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