Answer:
Velocity of jet in still air is 970 miles per hour and velocity of wind is 210 miles per hour.
Step-by-step explanation:
Jet's velocity against wind is
3040/4 = 760 miles per hour and flying with wind it is
8260/7 = 1180 miles per hour.
Let the velocity of jet in still air be x miles per hour and velocity of wind be y miles per hour.
As such its velocity against wind is x − y and with wind is x + y and therefore
x − y = 760 and x + y = 1180
Adding the two 2 x = 1940 and x = 970 and y = 1180 − 970 = 210
Hence velocity of jet in still air is 970 miles per hour and velocity of wind is
210 miles per hour.
Actual Answer:
https://socratic.org/questions/flying-against-the-wind-a-jet-travels-3040-miles-in-4-hours-flying-with-the-wind
Answer:
I think this would be....
14 miles im not sure
Step-by-step explanation:
I will assume that we are trying to find the slope between (-9,2) and (9, 2)
The slope's equation ⇒ 
<u>Let's set the variables</u>:
(x1, y1) --> (-9, 2)
(x2, y2) --> (9, 2)
<u>Now let's plug them in:</u>
<u>Thus the slope of the line is 0.</u>
Hope that helps!
Answer:
The answer is: y−9=2(x−5)
I took the quiz and the other answer here wasn't an answer choice but it said that this one was.
Answer: For problem 8:
sin A/a sinB/b = sinC/c
so sin50/15 = sinB/12
0.766/15 =sinB/12
0.051066 = sin B/12
sin B = 0.6128
B = 37.79 degrees
sum of angles must equal 180 deg therefore C = 180-50-37.79 = 92.21 degrees and .766/15 = sin92.21/c
.051066= 0.99925/c
c = .99925/0.051066 = 19.567
Problems 9 and 10 can also be done with the same method as problem 8.For problem 8:
sin A/a sinB/b = sinC/c
so sin50/15 = sinB/12
0.766/15 =sinB/12
0.051066 = sin B/12
sin B = 0.6128
B = 37.79 degrees
sum of angles must equal 180 deg therefore C = 180-50-37.79 = 92.21 degrees and .766/15 = sin92.21/c
.051066= 0.99925/c
c = .99925/0.051066 = 19.567
Problems 9 and 10 can also be done with the same method as problem 8.
<em>This is not my answer. I shall give credit to the rightful owner of these answers.</em>
