The speed of an airplane is 275 mi/h relative to the air. The wind is blowing due north with a speed of 25 mi/h. In what direction should the airplane head in order to arrive at a point due west of its location?

Step-by-step explanation:

The free-body diagram of the information given in the question is attached in the image below.

Using the property of Vector Right Triangle:

cos θ = adjacent/hypotenuse

cos θ = 25 mi/h / 275 mi/h

cos θ = 0.0909

θ = cos⁻¹ (0.0909)

θ = 84.79°

Therefore,  the airplane should head in the direction of S 84.79° W to reach the desired location.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3