Answer:
The side closest to P is the side that is on the same side of the angle bisector as P.
Step-by-step explanation:
The angle bisector is the line containing all the points equidistant from the sides of the angle. Points on one side of the angle bisector are closer to the angle side that is on that side of the angle bisector.
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The attached diagram shows the angle bisector as a dashed line. A couple of different locations for P are shown (P1 and P2). Apparently, we're concerned here with the distance from P along the perpendicular to each side of the angle. For P2 (on the left side of the angle bisector), it may be clear that the left perpendicular is shorter than the right one. Likewise, for P1, the right perpendicular will be shorter.
Answer:
15
Step-by-step explanation:
first you need to find the LCM (least common multiple) which is 15 so then every 15 days she will have both soccer, dance class and swimming class on the same day
Answer:
A 1/2
Step-by-step explanation:
Ratio of short length to hypotenuse
= cos60
= 1/2
Answer:
(in order) #1
for the first one you have a vertical angle. Vertical angle is basically any 2 lines who intersect each other. The opposite sides will be congruent.
So in this case
2x+8 and 110 are oppostie sides.
2x+8=110
Which is x= 51
#2
For the second one you have supplementray angles
Now for supplementary angles its basically 2 angles added up that form a straight line And 180 degress is always the measure of a straight line. SO in this case we do
9x+12x+75=180
Which is x=5
#3
This is also a vertical angle pair and their congruent just like #1
So 5x+5=65
x=12
Answer:
It would be option two: Bill's Electric Company charges 6 dollars per hour plus a flat rate of 15 dollars for a service call. If Monica has at least 100 dollars to spend, how many hours can the service man work?
Step-by-step explanation:
6 dollars an hour means it would be 6x.
A flat rate of 15 means there would be just 15 as a number, no x's.
And the sign is less than or equal to, so she has over a hundred dollars to spend.
