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Montano1993 [528]
3 years ago
7

Line m and point P are shown below. Part A: Using a compass and straightedge, construct line n parallel to line m and passing th

rough point P. Leave all construction marks. Part B: Explain the process that you used to construct line n.

Mathematics
1 answer:
Arturiano [62]3 years ago
5 0

Answer:

  see the attached

Step-by-step explanation:

In the attachment, we will refer to the circles, bottom-to-top, as circles 1, 2 and 3. The black points of intersection, bottom-to-top, will be referred to by the letters A, B, C, D. The transversal line through the white point (W) and pink point (P) will be line q.

Step 1. Draw line q through point P so it intersects line m at some convenient point. Label that point W.

Step 2. Choose an arbitrary radius for your compass. Here, we have chosen it to be the length WB. It happens to be less than half the length of WP, but that is not a requirement.

Step 3. Draw an arc of the chosen radius centered at W and intersecting line q and line m. Label the intersection points A (on line m) and B (on line q). These intersection points are on circle 1.

Step 4. Draw an arc of the same radius centered at P. It should be a long enough arc that it would intersect the proposed line parallel to m. Label the intersection point on line q with label C. This intersection point is on circle 3.

Step 5. Adjust the compass width (radius) to the distance from A to B. This is the radius of circle 2.

Step 6. Draw an arc centered at C so that it intersects the arc of Step 4. This is circle 2, and you want it to intersect circle 3. Label that point of intersection D.

Step 7. Draw line PD parallel to m.

_____

The point of the construction is to create congruent alternate interior angles AWB and CPD, so that lines AW and PD are parallel.

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Classify each number according to its value.
Tems11 [23]

Answer:See below and attached

Step-by-step explanation:

Given numbers

4.2×10^-6, 2.1×10^-3, 3.1×10^-2, 3.2×10^-5, 3.5×10^-4, 5.8×10^-3, 5.2×10^-4

Greater than 3.1×10^-3

3.1×10^-2, 5.8×10^-3

Between 3.1 × 10^-3  and 4.3 × 10^-5

2.1×10 ^-3, 3.5×10^-4, 5.2×10^-4

Less than 4.3 × 10^-5

4.2×10^-6, 3.2×10^-5

Step-by-step explanation

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3 years ago
at a small school, one male and one female are selected to represent the student body at the PTSCA meetings. if there are 41 mal
I am Lyosha [343]

It is given that there are 41 males and 48 females in the small school.

So, the number of ways a male student can be chosen from 41 males is ^{41}C_1

Likewise, the number of ways a female student can be chosen from 48 females is ^{48}C_1.

Thus, the total number of ways in which 2-person combinations are possible to represent the student body at the PTSAC meetings will be given by:

^{41}C_1\times ^{48}C_1=1968


7 0
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The length of a rectangle is four times its width.
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Let width be x

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ATQ

\\ \sf\longmapsto 2(x+4x)=70

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\\ \sf\longmapsto x=7

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2 years ago
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The answer is -1/4 I’m in 11th grade trust me

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With that 20 we can divide it by 180 to find her hourly rate

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Step-by-step explanation:

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