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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]\( a \)[/tex] and [tex]\( b \)[/tex] to create the equation that describes this situation.

[tex]\[ a \cdot x + b \quad x - b \][/tex]

Sagot :

GinnyAnswer
Let's solve this step by step.

We are given:

- The distance for each leg of the trip is 350 miles.
- The total trip time is 6 hours.
- The wind speed is 20 miles per hour.
- The speed of the airplane in still air is x.

First, let's write down the key points:

1. Speed of the airplane with the wind:

[tex]\[ \text{Speed with wind} = x + 20 \][/tex]

2. Speed of the airplane against the wind:

[tex]\[ \text{Speed against wind} = x - 20 \][/tex]

Next, let's define the time taken for each part of the trip. The time to travel a distance can be found using the formula:

[tex]\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \][/tex]

3. Time to travel to the destination with the wind:

[tex]\[ \text{Time to destination} = \frac{350}{x + 20} \][/tex]

4. Time to return against the wind:

[tex]\[ \text{Time back} = \frac{350}{x - 20} \][/tex]

The total time for the round trip is the sum of these two times and is given as 6 hours:

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

Now, let’s substitute the values in the expression:

[tex]\[ a = 350 \][/tex]
[tex]\[ b = 20 \][/tex]
[tex]\[ c = 6 \][/tex]

Thus, the equation that describes this situation is:

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

Or more specifically:

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

Here, the values are:
[tex]\[ a = 350, b = 20, c = 6 \][/tex]

So, replacing [tex]\( a \)[/tex] and [tex]\( b \)[/tex] in the given equation format [tex]\( \frac{a}{x+b} \quad \frac{a}{x-b} \)[/tex]:

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

Therefore, the corrected equation is

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