Answered

Join the IDNLearn.com community and start finding the answers you need today. Join our community to receive prompt, thorough responses from knowledgeable experts.

Enter the correct answer in the box.

An airplane makes a round-trip supply run that takes a total of 6 hours and is 350 miles each direction. The air current going to the destination is in the direction of the plane at 20 miles per hour. The air current traveling back to the starting point is against the plane at 20 miles per hour. Let [tex]$x$[/tex] represent the speed of the airplane, in miles per hour, when there is no wind.

Replace the values of [tex]$a, b$[/tex] and [tex][tex]$c$[/tex][/tex] to create the equation that describes this situation.

[tex]
\frac{a}{x+b}+\frac{a}{x-b}=c
[/tex]

Sagot :

GinnyAnsweridn
To describe the given situation with the appropriate values substituted into the equation, you start by understanding the basic form of the equation:

[tex]\[ \frac{a}{x+b} + \frac{a}{x-b} = c \][/tex]

In this equation:
- [tex]\( a \)[/tex] is the one-way distance,
- [tex]\( b \)[/tex] is the wind speed,
- [tex]\( c \)[/tex] is the total time for the round trip.

Given the values:
- [tex]\( a = 350 \)[/tex] miles,
- [tex]\( b = 20 \)[/tex] miles per hour,
- [tex]\( c = 6 \)[/tex] hours,

we substitute these values into the equation, resulting in:

[tex]\[ \frac{350}{x+20} + \frac{350}{x-20} = 6 \][/tex]

This equation accurately models the problem where [tex]\( x \)[/tex] represents the speed of the airplane in still air, and considers the influence of the wind speed in each direction of the round trip.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Thank you for choosing IDNLearn.com for your queries. We’re here to provide accurate answers, so visit us again soon.