Answer:
cosФ = 
, sinФ = 
, tanФ = -8, secФ = 
, cscФ = 
, cotФ = 
Step-by-step explanation:
If a point (x, y) lies on the terminal side of angle Ф in standard position, then the six trigonometry functions are:
- cosФ =
- sinФ =
- tanФ =
- secФ =
- cscФ =
- cotФ =
- Where r =
(the length of the terminal side from the origin to point (x, y)
- You should find the quadrant of (x, y) to adjust the sign of each function
∵ Point (1, -8) lies on the terminal side of angle Ф in standard position
∵ x is positive and y is negative
→ That means the point lies on the 4th quadrant
∴ Angle Ф is on the 4th quadrant
∵ In the 4th quadrant cosФ and secФ only have positive values
∴ sinФ, secФ, tanФ, and cotФ have negative values
→ let us find r
∵ r =
∵ x = 1 and y = -8
∴ r = 
→ Use the rules above to find the six trigonometric functions of Ф
∵ cosФ =
∴ cosФ =
∵ sinФ =
∴ sinФ =
∵ tanФ =
∴ tanФ = 
= -8
∵ secФ =
∴ secФ = 
=
∵ cscФ =
∴ cscФ =
∵ cotФ =
∴ cotФ =
Answer:
A and C
Step-by-step explanation:
A: 5 tens = 50
9 tenths = 0.9
6 hundreths = 0.06
= 50.96
C: 5*10 = 50 + 9 * 1/10 = 0.9 + 6 * 1/100 = 50.96
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^2 - x = 7
(x - 0.5)^2 - 0.25 = 7
(x - 0.5)^2 = 7.25
(x - 0.5) = +/- sqrt 7.25
x = 0.5 +/- sqrt7.25
solution set is {-2.19, 3.19,}
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