PLEASE HELP A flight across the US takes longer east to west then it does west to east. This is due to the plane having a headwind flying westward (thus slowing it down) and a tailwind flying eastward (thus pushing it along faster). Suppose you are flying from Philadelphia, PA to San Diego, CA. The distance between the cities is 2,400 miles and the plane's speed is 450 mi/hr. What is the speed of the wind?Complete the chartGoing westward- Distance - Rate- TimeGoing Eastward-Write an equation to represent the total time if the round trip takes 11 hours.Solve your equation to find the speed of the wind. Round to the nearest mph. PLEASE HELP!PLEASE ANSWER CORRECTLY TO GET BRAINLIEST ANSWER

To solve our questions, we are going to use the kinematic equation for distance: [tex]d=vt[/tex]
where
[tex]d[/tex] is distance 
[tex]v[/tex] is speed  
[tex]t[/tex] is time 

1. Let [tex]v_{w}[/tex] be the speed of the wind, [tex]t_{w}[/tex] be time of the westward trip, and [tex]t_{e}[/tex] the time of the eastward trip. We know from our problem that the distance between the cities is 2,400 miles, so [tex]d=2400[/tex]. We also know that the speed of the plane is 450 mi/hr, so [tex]v=450[/tex]. Now we can use our equation the relate the unknown quantities with the quantities that we know:

Going westward:
The plane is flying against the wind, so we need to subtract the speed of the wind form the speed of the plane:
[tex]d=vt[/tex]
[tex]2400=(450-v_{w})t_{w}[/tex]

Going eastward:
The plane is flying with the wind, so we need to add the speed of the wind to the speed of the plane:
[tex]d=vt[/tex]
[tex]2400=(450+v_{w})t_{e}[/tex]

We can conclude that you should complete the chart as follows:
Going westward -Distance: 2400 Rate:[tex]450-v_w[/tex] Time:[tex]t_w[/tex]
Going eastward -Distance: 2400 Rate:[tex]450+v_w[/tex] Time:[tex]t_e[/tex]

2. Notice that we already have to equations:
Going westward: [tex]2400=(450-v_{w})t_{w}[/tex] equation(1)
Going eastward: [tex]2400=(450+v_{w})t_{e}[/tex] equation (2)

Let [tex]t_{t}[/tex] be the time of the round trip. We know from our problem that the round trip takes 11 hours, so [tex]t_{t}=11[/tex], but we also know that the time round trip is the time of the westward trip plus the time of the eastward trip, so [tex]t_{t}=t_w+t_e [/tex]. Using this equation we can express [tex]t_w[/tex] in terms of [tex]t_e[/tex]:
[tex]t_{t}=t_w+t_e [/tex]
[tex]11=t_w+t_e[/tex] equation
[tex]t_w=11-t_e[/tex] equation (3)
Now, we can replace equation (3) in equation (1) to create a system of equations with two unknowns: 
[tex]2400=(450-v_{w})t_{w}[/tex]
[tex]2400=(450-v_{w})(11-t_e)[/tex] 

We can conclude that the system of equations that represent the situation if the round trip takes 11 hours is:
[tex]2400=(450-v_{w})(11-t_e)[/tex] equation (1)
[tex]2400=(450+v_{w})t_{e}[/tex] equation (2)

3. Lets solve our system of equations to find the speed of the wind: 
[tex]2400=(450-v_{w})(11-t_e)[/tex] equation (1)
[tex]2400=(450+v_{w})t_{e}[/tex] equation (2)

Step 1. Solve for [tex]t_{e}[/tex] in equation (2)
[tex]2400=(450+v_{w})t_{e}[/tex]
[tex]t_{e}= \frac{2400}{450+v_{w}} [/tex] equation (3)

Step 2. Replace equation (3) in equation (1) and solve for [tex]v_w[/tex]:
[tex]2400=(450-v_{w})(11-t_e)[/tex]
[tex]2400=(450-v_{w})(11-\frac{2400}{450+v_{w}} )[/tex]
[tex]2400=(450-v_{w})( \frac{4950+11v_w-2400}{450+v_{w}} )[/tex]
[tex]2400=(450-v_{w})( \frac{255011v_w}{450+v_{w}} ) [/tex]
[tex]2400= \frac{1147500+4950v_w-2550v_w-11(v_w)^2}{450+v_{w}} [/tex]
[tex]2400(450+v_{w})=1147500+2400v_w-11(v_w)^2 [/tex]
[tex]1080000+2400v_w=1147500+2400v_w-11(v_w)^2[/tex]
[tex](11v_w)^2-67500=0[/tex]
[tex]11(v_w)^2=67500[/tex]
[tex](v_w)^2= \frac{67500}{11} [/tex]
[tex]v_w= \sqrt{\frac{67500}{11}} [/tex]
[tex]v_w=78[/tex]

We can conclude that the speed of the wind is 78 mi/hr.