Los Angeles Airport bears 140 degrees from San Francisco Airport and is 540 km away. A pilot is planning a direct flight from San Francisco Airport to Los Angeles Airport to leave at 2 P.M. The plane's air speed will be 640 km/h, and there will be a 60 km/hwind blowing from 290 degrees. What should the compass heading be,and what is the plane's estimated time of arrival (ETA) to the nearest minute?
Answers
Answer 1
Answer:
Create a triangle starting with the length of 540 coming out of SF to LA bearing 140.
This looks like it leaves SF in IV quadrant and enters LA in II quadrant. The wind will take the plane more to the east (right). Therefore the pilot must aim to the west (left) in order for the wind to push it to the intended destination of LA. The second side of the triangle should come out of LA from the II quadrant bearing 290. The length of this side is 60t (60 km/h * t = time flying). The third side of the triangle connects the first two. There for it comes out of SF at some unknown angle > 140 and connects with the 60t side. The length of this third leg is 640t. (640 km/h * t = time flying). The angle between the 540 side and the 60t side is 30. This is found because the 540 side enters LA in 2nd quadrant as 130 angle . The 60t side enters LA in 2nd quadrant as 160 angle. Using law of sin's: sin 30/640t = sin x/60t. The t's cancel and you are left with sin x = 3/64. When solving for the angle x you get x = 2.6867 degrees. Adding this to the bearing of 140, the compass bearing should be 142.6867 or 142.7 degrees. To find the value of t, you use the law of sin's to get sin 30/640t = (sin (180 - 30 - 2.6867))/540. Solving for t gives you: t = .7811. This is in hours. To convert to minutes multiply by 60 to get t = 46.87 minutes. Add this to the 2pm departure time to get 2:47 pm arrival time.
This looks like it leaves SF in IV quadrant and enters LA in II quadrant. The wind will take the plane more to the east (right). Therefore the pilot must aim to the west (left) in order for the wind to push it to the intended destination of LA. The second side of the triangle should come out of LA from the II quadrant bearing 290. The length of this side is 60t (60 km/h * t = time flying). The third side of the triangle connects the first two. There for it comes out of SF at some unknown angle > 140 and connects with the 60t side. The length of this third leg is 640t. (640 km/h * t = time flying). The angle between the 540 side and the 60t side is 30. This is found because the 540 side enters LA in 2nd quadrant as 130 angle . The 60t side enters LA in 2nd quadrant as 160 angle. Using law of sin's: sin 30/640t = sin x/60t. The t's cancel and you are left with sin x = 3/64. When solving for the angle x you get x = 2.6867 degrees. Adding this to the bearing of 140, the compass bearing should be 142.6867 or 142.7 degrees. To find the value of t, you use the law of sin's to get sin 30/640t = (sin (180 - 30 - 2.6867))/540. Solving for t gives you: t = .7811. This is in hours. To convert to minutes multiply by 60 to get t = 46.87 minutes. Add this to the 2pm departure time to get 2:47 pm arrival time.
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Answers
Four friends were born on consecutive years.
The sum of their ages is 62. Let's find the age of the youngest friend:
Formula:
4x + 6 = 62
=> First is to Subtract 6 from both sides 4x = 56
Then next thing to do is to divide 4 from both sides:
=> x = 14 14 + 14 + 1 + 14 + 2 + 14 + 3 = 62
62 = 62
Thus the youngest age is 14
The sum of their ages is 62. Let's find the age of the youngest friend:
Formula:
4x + 6 = 62
=> First is to Subtract 6 from both sides 4x = 56
Then next thing to do is to divide 4 from both sides:
=> x = 14 14 + 14 + 1 + 14 + 2 + 14 + 3 = 62
62 = 62
Thus the youngest age is 14
Answers
The subject, so that the recipient knows what the email is about but only through a couple words
Answers
Let x= side 1= 19
Let y= side 2=unknown
Let z= side 3=unknown
Since it is an isosceles triangle, two sides are equal, which happen to be y and z
Hence y=z
Perimeter of a triangle = Sum of all three sides
= x+y+z
51=19+2y [since we know x=19 and y=z, we know the perimeter is 51]
Solve now
32=2y
16=y
Since y=z
z= 16
Hence the unknown side's length is 16m
Let y= side 2=unknown
Let z= side 3=unknown
Since it is an isosceles triangle, two sides are equal, which happen to be y and z
Hence y=z
Perimeter of a triangle = Sum of all three sides
= x+y+z
51=19+2y [since we know x=19 and y=z, we know the perimeter is 51]
Solve now
32=2y
16=y
Since y=z
z= 16
Hence the unknown side's length is 16m
So the perimeter is 51. The base is 19. Subtract 19 from 51. 51 - 19 = 32. So 32 is the total length of the two equal unknown sides added together. Because the two sides must be equal, you can just divide 32 by 2. 32/2 = 16. Each of the two congruent sides are 16 m.
Answers
dunnow what marthian
percent means parts ot of 100
15%=15/100=0.15
'of' means multpily
15% of 60 means
0.15 times 60=9
answer=9
percent means parts ot of 100
15%=15/100=0.15
'of' means multpily
15% of 60 means
0.15 times 60=9
answer=9
15% of 60
15/100 = .15
.15*60 = 9
Answer: 9
15/100 = .15
.15*60 = 9
Answer: 9
Answers
the answer to your question is (8(2a+9)
Answers
OK.......
So,
ON MONDAY:
Time spent on patrol = 4.5 hrs
Time spent on surveillance = 2 hrs
ON TUESDAY:
Time spent on patrol = 3 hrs
Time spent on surveillance = 1 hr 40 min
Now,
The question is asking us the ratio of surveillance to patrol so, the value of surveillance needs to be on the top.
So,
Let's first change all the units to common units.....i.e. either hrs. to minutes or minutes to hrs.
In this case .....i think it will be easier to convert minutes to hrs.. because if we do that ...we only have to do one conversion.
So,
1 hr. 40 minuter = 1hr + (40 /60) hrs
= 1 + 0.67
= 1.67 hrs
So,
ON TUESDAY:
Time spent on surveillance = 1.67 hrs
Now,
Let us find the ratio of surveillance to patrol on both days:
ON MONDAY:
Ratio of surveillance to patrol =
ON TUESDAY:
Ratio of surveillance to patrol =
Now,
If we look at the ratio, the ratio of TUESDAY sure is greater than ratio on MONDAY
So, Tuesday has a greater ratio of time on surveillance to time on patrol
So,
ON MONDAY:
Time spent on patrol = 4.5 hrs
Time spent on surveillance = 2 hrs
ON TUESDAY:
Time spent on patrol = 3 hrs
Time spent on surveillance = 1 hr 40 min
Now,
The question is asking us the ratio of surveillance to patrol so, the value of surveillance needs to be on the top.
So,
Let's first change all the units to common units.....i.e. either hrs. to minutes or minutes to hrs.
In this case .....i think it will be easier to convert minutes to hrs.. because if we do that ...we only have to do one conversion.
So,
1 hr. 40 minuter = 1hr + (40 /60) hrs
= 1 + 0.67
= 1.67 hrs
So,
ON TUESDAY:
Time spent on surveillance = 1.67 hrs
Now,
Let us find the ratio of surveillance to patrol on both days:
ON MONDAY:
Ratio of surveillance to patrol =
ON TUESDAY:
Ratio of surveillance to patrol =
Now,
If we look at the ratio, the ratio of TUESDAY sure is greater than ratio on MONDAY
So, Tuesday has a greater ratio of time on surveillance to time on patrol
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