So this is taking this equation and plugging it into the quadratic formula -b+- (sqrt (b^2)-(4)ac)/2a so your A value is 1 B value is 3 and C value is -5 so it’s B&E
Answer:
x = 800 km/h
y = 200 km/h
the rate of the plane in still air is 800 km/h
and the rate of the wind is 200 km/h
Completed question;
Flying against the wind, an airplane travels 4200 km in 7 hours. Flying with the wind, the same plane travels 4000 km in 4 hours. What is the rate of the plane in still air and what is the rate of the wind?
Step-by-step explanation:
Let x and y represent the rate of the plane and wind respectively;
When flying against the wind, the relative speed is;
Va = x-y .......1
When flying with the wind, the relative speed is;
Vb = x+y .......2
Distance = speed × time
Speed = distance/time
v = d/t
Given;
Flying against the wind, an airplane travels 4200 km in 7 hours. Flying with the wind, the same plane travels 4000 km in 4 hours
When flying against wind;
distance da = 4200 km
time ta = 7 hours
Va = da/ta = 4200/7
Va = 600 km/h
Substituting equation 1
x-y = 600 .....3
When flying with wind;
distance db = 4000 km
Time tb = 4 hours
Vb = db/tb = 4000/4
Vb = 1000 km/h
Substituting equation 2;
x + y = 1000 .....4
Adding equation 3 and 4;
x-y + (x+y) = 600 + 1000
2x = 1600
x = 1600/2
x = 800 km/h
Substituting x = 800 into equation 3;
800 - y = 600
y = 800 - 600
y = 200 km/h
Therefore, the rate of the plane in still air is 800 km/h
and the rate of the wind is 200 km/h.
Answer: (2,8), because the graph of the two equations intersects at this point.
Step-by-step explanation: Got u. it's right 1000%
2. line segment MX is congruent to line segment NX
4. All right angles are congruent
5. line segment QX is congruent to line segment QX
6. ASA Congruence Postulate
7. SAS Congruence Postulate
Hope this helps