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sweet [91]
3 years ago
14

write an equation that is perpendicular to x - 3y =3 and passes through the point (5,-9) in slope-intercept form.

Mathematics
1 answer:
Aleksandr [31]3 years ago
6 0

9514 1404 393

Answer:

  y = -3x +6

Step-by-step explanation:

The perpendicular line will have the coefficients of x and y swapped (and one of them negated), and will have a constant appropriate to the point the line needs to go through.

The equation will look like ...

  x -3y = 3 . . . . . . . . . . . . . given line

  3x +y = constant . . . . . . perpendicular line

where "constant" can be found by substituting for x and y.

  3x +y = 3(5) +(-9)

  3x +y = 6 . . . . equation in standard form

To put this in slope-intercept form subtract 3x:

  y = -3x +6

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3 years ago
Find 5 solutions to the question.<br><br> Y=-1.5x+2
Black_prince [1.1K]

Answer:

<em>☆</em><em><</em><em> </em><em><u>《</u></em><em><u>HOPE IT WILL HELP YOU</u></em><em>》</em><em>></em><em>☆</em>

Step-by-step explanation:

Y=1.5x+2

<h3><em><u>(</u></em><em><u>x</u></em><em><u>=</u></em><em><u>1</u></em><em><u>)</u></em></h3>

y=1.5 (1) +2

y=1.5+2

y=3.5

<h3><em><u>(</u></em><em><u>x</u></em><em><u>=</u></em><em><u>2</u></em><em><u>)</u></em></h3>

y=1.5(2)+2

y=3+2

y=5

<h3><em><u>(</u></em><em><u>x</u></em><em><u>=</u></em><em><u>3</u></em><em><u>)</u></em></h3>

y=1.5 (3)+2

y=4.5+2

y=6.5

<h3>(x=4)</h3>

y=1.5 (4)+2

y=6+2

y=8

<h3><em><u>(</u></em><em><u>x</u></em><em><u>=</u></em><em><u>5</u></em><em><u>)</u></em></h3>

y=1.5 (5)+2

y=7.5+2

y=9.5

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How do you solve and check x/4 = -5
Mashutka [201]

Rearrange x/4-(-5)=0

Simplify x/4

-5 = -5/1 = -5*4/4

x- (-5*4)/4 = x+20/4

x+ 20/4 = 0

x+20/4 * 4 = 0*4

The equation now takes the shape: x+20=0

subtract 20 from both sides of the equation: x=-20

Answer: x=-20

6 0
3 years ago
7 divided by 4 x + 3 Y + 4 divided / 4 x minus 3 Y is equal to 5 / 4
DerKrebs [107]

9514 1404 393

Answer:

  (x, y) = (4, 4)

Step-by-step explanation:

If you attempt to solve this by clearing fractions, you end up with an extraneous solution. Here, we'll solve the linear equations ...

  7a +4b = 5/4

  8b -14a = 3/2

where a = 1/(4x+3y) and b = 1/(4x-3y)

Dividing the second equation by 2 and adding the first, we have ...

  (7a +4b) +1/2(8b -14a) = (5/4) +1/2(3/2)

  8b = 8/4

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Substituting into the first equation gives ...

  7a +4/4 = 5/4

  a = 1/28

__

Now, we can get back to solving for x and y.

  4x +3y = 1/a = 28

  4x -3y = 1/b = 4

  8x = 32 . . . . . . . . . add the two equations

  x = 4

  3y = 28 -4x = 28 -4(4) = 12 . . . . . use x in the equation for 'a'; rearrange

  y = 4 . . . . divide by 3

The solution to the system of equations is (x, y) = (4, 4).

_____

<em>Additional comment</em>

If you graph these equations, you find they describe hyperbolas that intersect at (0, 0) and (4, 4). The "solution" (0, 0) is extraneous, as both equations are undefined there.

7 0
2 years ago
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