Let [tex] v_p [/tex] be the speed of the plane in still air. Let [tex] t [/tex] be the time taken by the plane with the wind. Then, from the given information,
[tex] (v_p+50)t=551\\
(v_p-50)(t+1)=252\\ [/tex]
From the above 2 equations,
[tex] (v_p+50)t=551\\
(v_p-50)t=302-v_p\\ [/tex]
Dividing the above 2 equations,
[tex] \frac{v_p-50}{v_p+50} =\frac{302-v_p}{551} \\
551(v_p-50)=(302-v_p)(v_p+50)\\
v_p^2+299v_p-42650=0\\
v_p=105.45
[/tex]
The speed of the plane in still air is [tex] 105.45 \;mi/hour [/tex].
Answer:
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Explanation:
In photo below
I hope this helps
