A hot- air balloon is rising from the ground at a rate of 4 feet every second. Another balloon at 756 feet is descending at a rate of 3 feet every second. In how many seconds will the two balloons be at the same altitude.

Respuesta :

Answer:

108 seconds.

Step-by-step explanation:

Let x represent number of seconds.

We have been given that a hot- air balloon is rising from the ground at a rate of 4 feet every second. So altitude of hot-air balloon would be [tex]4x[/tex] feet after x seconds.

We are also told that another balloon at 756 feet is descending at a rate of 3 feet every second. Since the altitude of 2nd balloon is descending, so altitude after x seconds would be initial altitude minus change in altitude after x seconds [tex]756-3x[/tex].

Now, we will equate both expression to find the time as:

[tex]4x=756-3x[/tex]

[tex]4x+3x=756-3x+3x[/tex]

[tex]7x=756[/tex]

[tex]\frac{7x}{7}=\frac{756}{7}[/tex]

[tex]x=108[/tex]

Therefore, the two balloons will be at the same altitude in 108 seconds.