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.
X/-9 ≥3 multiply both side by -9 to get rid of fraction
-9/1•x/-9 ≥-9•3
Cross cancel -9 on left side
x ≥ -27
Answer: 
Step-by-step explanation:
Given : Two opposite sides of a rectangle are each of length x.
Let the other adjacent side be y.
The perimeter of the rectangle is 12 units.
Perimeter of rectangle is given by :-
The area of rectangle is given by :-
Hence, the area as a function x =
Answer:
2,160 seconds
Step-by-step explanation:
Answer:
A. -12
Step-by-step explanation:
A graph shows the vertices of the feasible region to be (0, 6), (3, 0) and (0, -3). Of these, the one that minimizes f(x, y) is (0, -3). The minimum value is ...
f(0, -3) = 3·0 + 4(-3) = -12
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<em>Comment on the graph</em>
Here, three regions overlap to form the region where solutions are feasible. By reversing the inequality in each of the constraints, <em>the feasible region shows up on the graph as a white space</em>, making it easier to identify. The corner of the feasible region that minimizes the objective function is the one at the bottom, at (0, -3).
