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katovenus [111]
3 years ago
15

Please help i am stuck

Mathematics
1 answer:
balu736 [363]3 years ago
6 0

Explanation:

All of these are done the same way.

1. Set your compass to a distance slightly longer than the distance from the point to the line you want the perpendicular to.

2. Using the point as a center, draw an arc that intersects the line in 2 places. Consider those points to be "P" and "Q".

3. You can leave the compass as is, or set it to any other distance longer than half the distance between P and Q. Using this radius, and using P and Q as centers, draw intersecting arcs on the other side of the line from the original point. Consider the intersection of those arcs to be point "R".

4. A line through the original point and point R will be perpendicular to the line, as you want.

_____

In problem 1, the point and line are obvious.

In problem 2, the point is the vertex opposite the line of interest. (There will be 3 similar constructions.)

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How far is the portion of her route from A to D? Round to the nearest whole number. Explain. I NEED IT BY TONIGHT SO PLSS SOMEON
inn [45]

9514 1404 393

Answer:

  A to D is about 2087 ft

Step-by-step explanation:

For the portion of the path of interest, the sine and cosine relations apply.

  Sin = Opposite/Hypotenuse

  Cos = Adjacent/Hypotenuse

__

AB is the adjacent side to the angle marked 33°, with hypotenuse 975 ft.

  cos(33°) = AB/(975 ft)

  AB = (975 ft)cos(33°) ≈ 817.70 ft

BC is the opposite side in the same triangle.

  sin(33°) = BC/(975 ft)

  BC = (975 ft)sin(33°) ≈ 531.02 ft

CD is the hypotenuse of the right triangle BCD. The side BC that we know is opposite the given angle, so the sine relation applies.

  sin(46°) = BC/CD

  CD = BC/sin(46°)

  CD = (531.02 ft)/sin(46°) ≈ 738.21 ft

__

Now we know the segments of the path of interest:

  A–D = AB +BC +CD

  A–D = 817.70 ft + 531.02 ft + 738.21 ft = 2086.93 ft

  A–D ≈ 2087 ft

_____

<em>Additional comments</em>

The attachment shows all of the segments computed working counterclockwise from A. The angles in triangle DEF are computed using the Law of Sines from sides DF and DE and angle DEF. Segments EG and GH are computed using the Law of Cosines.

When we get to points F, G, H, we find that there is some inconsistency with the locations that would be computed working clockwise using AB and angle BFA. This inconsistency shows up in an error in the 119° angle in the attached figure.

This means that segments on the back side of the route, along path EFGHA, will vary somewhat depending on how they're computed.

We assume that points B, C, F, G are collinear.

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3 years ago
Sort the data from least to greatest.
gizmo_the_mogwai [7]
0 0 0 1 2 2 4 thats the answer
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3 years ago
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