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2 years ago

Determine if the given equations are parallel, perpendicular, or neither 4y=3x-4 and 4x+3y=-6

Mathematics

1 answer:

Kazeer [188]2 years ago

8 0

Answer:

Perpindicular

Step-by-step explanation:

Solve each equation for y, or point Slope form.

y = mx + b

4y = 3x - 4 divide both sides by 4 to isolate the y.

y = 3/4x - 1

The second equation.

4x + 3y = - 6 subtract 4x from both sides

3y = -4x - 6 divide both sides by 3

y = -4/3 - 2

If you visualize line 1 having a Slope of 3/4, that is, for every 4 we move in the positive x direction, y will increase by 3, it slopes up from left to right crossing the y axis at (0,-1). The second line moves 3 in the positive x direction and decreases in y by 4, it slopes down from left to right crossing the y axis at (0,-2}.

Because the two lines have negative inverse slopes they are perpendicular. 3/4x and - 4/3x

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2 years ago

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2 years ago

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2 years ago

Match the compound inequalities

kondor19780726 [428]

I one is 8x<24 and -8≤2x-4

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Hence solution is -2≤x<3

Therefore c is the correct matching for 1.

2) 5x-2>13 or -4x≥8

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Or x>3 or x ≤-2

Hence solution is two regions to the right of 3 excluding 3 and left of -2 including -2.

Graph b is the correct match.

3) -25≤9x+2<20

Subtract 2

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-3≤x<2

Hence graph is the region between -3 and 2 including only -3.

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3 years ago

The center of a hyperbola is (−4,3) , and one vertex is (−4,7) . The slope of one of the asymptotes is 2.

Monica [59]

Answer:

The answer to your question is below

Step-by-step explanation:

C (-4, 3)

V (-4, 7)

asymptotes = 2 = $\frac{b}{a}$

- This is a vertical hyperbola, the equation is

$\frac{(y - k)^{2} }{a^{2} } + \frac{(x - h)^{2} }{b^{2} } = 1$

slope = 2

a is the distance from the center to the vertex = 4

b = 2(4) = 8

$\frac{(y - 3)^{2} }{4^{2} } + \frac{(x + 4)^{2} }{8^{2} } = 1$

$\frac{(y - 3)^{2} }{16} + \frac{(x + 4)^{2} }{64} = 1$

7 0

3 years ago

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