Factor -9z + 12 A; -3(2z + 4) B: -9(z - 4) C: -9(z + 12) D: -3(3z - 4)
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Problem 1
Draw a straight line and plot P anywhere on it. Use the compass to trace out a faint circle of radius 8 cm with center P. This circle crosses the previous line at point Q.
Repeat these steps to set up another circle centered at Q and keep the radius the same. The two circles cross at two locations. Let's mark one of those locations point X. From here, we could connect points X, P, Q to form an equilateral triangle. However, we only want the 60 degree angle from it.
With P as the center, draw another circle with radius 7.5 cm. This circle will cross the ray PX at location R.
Refer to the diagram below.
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Problem 2
I'm not sure why your teacher wants you to use a compass and straightedge to construct an 80 degree angle. Such a task is not possible. The proof is lengthy but look up the term "constructible angles" and you'll find that only angles of the form 3n are possible to make with compass/straight edge.
In other words, you can only do multiples of 3. Unfortunately 80 is not a multiple of 3. I used GeoGebra to create the image below, as well as problem 1.
Given:
The figure of two parallel lines MP and QS.
A transversal line KL interests the two parallel lines.
To find:
The .
Solution:
If a transversal line intersect the two parallel lines, then the corresponding angles are congruent and their measures are equal.
In the given figure and
corresponding angles. It means their measures are equal.
Therefore, .