HELP ME I NEED PART A. And PART B SO PLZ HELP ME I DONT GET IT

The graph of 3x - 2y = 6 does not pass through? (4,-3) (4/3,-1) (-2,-6)

victus00 [196]

(4,-3) does not pass through the graph

If you put (4, -3) into the equation you get 18

If you put (4/3, -1) into the equation you get 6

If you put (-2, -6) into the equation you get 6

3 0

2 years ago

Algebra: Explain in words how to write an equation that is part one: parallel and then also part two: perpendicular to the equat

sweet-ann [11.9K]

A line that is parallel has the same slop where a line that is perpendicular has a slop that is negative and reciprocal.

so for the parallel one you don't need to worry about the slop because it will be 2/3x. But yous the point slope equation form

y-y1=m(x-x1)

y+5=2/3(x+2)

y+5=2/3x+ 4/3

y=2/3x-11/3

-2/3x+y=-11/3

multiple by -1 so A inst negative

2/3-y=11/3

For a line that is perpendicular you just need to flip the original 2/3x slope and make it negative.

y+5=-3/2(x+2)

y+5=-3/2x-3

y=-3/2x-8

3/2x+y=-8

8 0

2 years ago

Solve the given triangles by finding the missing angle and other side lengths.

coldgirl [10]

Answer:

a ≈ 27.65; B = 28°; c ≈ 19.17
a ≈ 163.30; B = 120°; c ≈ 59.77
A = 13°; b ≈ 0.3874; c ≈ 1.3737

Step-by-step explanation:

As long as you're seeking outside help, the use of a suitable tool is appropriate. Many graphing calculators, apps, and web sites are available for solving triangles.

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Here, you're given two angles and a side. The third angle is the difference from 180° of the sum of the other two. For example, in the first problem, ...

B = 180° -A -C = 28°

The missing sides are found using the law of sines. Since you are given a side, use that as the reference, and its opposite angle as the reference angle. Then you have ...

side = sin(opposite angle)×((reference side)/sin(reference angle))

Note that the factor in italics remains the same for both remaining sides.

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In the first problem, this becomes ...

a = sin(112°)·14/sin(28°) ≈ 27.65

c = sin(40°)·14/sin(28°) ≈ 19.17

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The remaining problems are worked in similar fashion.

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Comment on triangle solutions

These are straightforward because you're given two angles. If you're given two sides and an angle, the solution method varies depending on which angle you're given (opposite the long side, the short side, or the unknown side). In the last case, the law of cosines is involved. In the second case, there are likely two solutions. Once you have two angles, you can proceed as above.