this is the answer to this question - Two weather tracking stations are on the equator 146 miles apart. A weather balloon is located on a bearing of N 35°E from the western station and on a bearing of N 23°E from the eastern station. How far is the balloon from the western station?
Answer:
Reasons:
The given parameters are;
Distance between the two stations = 146 miles
Location of the weather balloon from the Western station = N35°E
Location of the weather balloon from the Eastern station = N23°E
The location of the station = On the equator
Required:
The distance of the balloon from the Western station
Solution:
- The angle formed between the horizontal, and the line from the Western station
to the balloon = 90° - 35° = 55°
- The angle formed between the horizontal, and the line from the Eastern station
to the balloon = 90° + 23° = 113°
The angle at the vertex of the triangle formed by the balloon and the two stations is 180° - (55 + 113)° = 12°
By sine rule,
Distance from balloon to western station = 146/sin(12 dg) = Distance from balloon to western station/sin(113 dg)
Therefore;
Distance from balloon to western station = 146/sin(12 dg) x sin(113 dg) ~ 646.4
Step-by-step explanation:
Answer:
120 - x - 9 past 10am
Step-by-step explanation:
Call “x” the number of minutes it is before 12 noon. Measure time in minutes from 10am, so that 12 noon is 2*60 = 120 minutes after 10 am. The current time is 120 - x after 10 am. Nine minutes ago it was 120 - x - 9 past 10am.
So the 74% that you have right now is the other 80% of your grade. So you multiply it by its weight and solve for the other one making it equal to 78%.
(.74)(.8) + (.2)(x) = .78
.592 + (.2)(x) = .78
(.2)(x) = .188
(x) = .94
So, you need to get a 94% on your final test.
Answer:
100 times more
Step-by-step explanation:
if you multiply 7,000,000 by 1000 you get 7,000,000,000
