Two hot air balloons are flying above a park. One balloon started at a height of 3,000 feet above the ground and is decreasing in height at a rate of 40 feet per minute. The second balloon is rising at a rate of 50 feet per minute after beginning from a height of 1,200 feet above the ground. Given that h is the height of the balloons after m minutes, determine which system of equations represents this situation. A. h = 3,000 - 40m h = 1,200 50m B. m = 3,000 - 40h m = 1,200 50h C. h = 3,000 40m h = 1,200 - 50m D. h = 3,000m - 40 h = 1,200m 50

where m is the slope or rate of change--can be positive or negative--and b is the y-intercept or the initial value of the function when x = 0.

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Breaking down the Problem

One balloon starts at an altitude of 3,000 feet before the first minute struck, this is our b value. It then decreases by 40 feet every minute, so -40 is our slope value.

So, the equation for the first balloon is

y = -40x + 3000,

or

h = -40m + 3000,

where m is the number of minutes that pass--no longer the rate of change--, and h is the elevation after m minutes.

The second balloon starts at an altitude of 1,200 feet before the first minute struck, this is our b value. Its altitude increases by 50 feet every minute, so 50 is our slope value.

So, the equation for the second balloon is

y = 50x + 1200,

or

h = 50m + 1200.

Both equations of this system match answer choice A, so that's our answer!