Answer is <span>B.
Quadrant II </span>
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or …show more content…
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or half-moon status because at least 50 percent of its surface was illuminated. In the following nights, the Moon displayed characteristics of waxing gibbous as the light continued to grow across the moon’s surface from right to left. The Moon was nearing closer to the full moon phase on November 14th as only a very small dark shadow was visible on the left side.
The Moon takes 27.3 days (sidereal month) to complete its actual orbit around the Earth. Like the Sun, the Moon rises in the east and sets in the west each day.
You can first take out an x^2 from each number , 3x^6 - 39x^4 + 108x^2 = x^2 ( 3x^4 - 39x^2 + 108) Then just factor and solve for the zero's of the equation
Answer:
6
Step-by-step explanation:
We know that MY is the same length as XM according to definition of segment bisector. We can create an equation:
5x + 8 = 9x + 12
Minus 8 on both sides
5x = 9x + 4
-4 = 4x
x = -1
Now substitute both sides of the equation with -1 instead of x
-5 + 8 = -9x + 12
3 = 3
XY = MY + XM
XY = 3 + 3 = 6
Hope this helps :)
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