1. **State the problem:** Two birds, A and B, are flying at different heights and changing their altitudes at constant rates. We need to find their height equations, when they are at the same height, and what that height is. 2. **Define variables and write equations:** Let $t$ be the time in minutes after they start changing altitude. - Bird A starts at 500 feet and descends at 75 feet per minute, so its height decreases by 75 each minute: $$h_A(t) = 500 - 75t$$ - Bird B starts at 100 feet and ascends at 25 feet per minute, so its height increases by 25 each minute: $$h_B(t) = 100 + 25t$$ 3. **Find when the birds are at the same height:** Set $h_A(t) = h_B(t)$: $$500 - 75t = 100 + 25t$$ 4. **Solve for $t$:** Move terms to one side: $$500 - 75t - 100 - 25t = 0$$ $$400 - 100t = 0$$ Isolate $t$: $$100t = 400$$ $$t = \frac{400}{100}$$ $$t = 4$$ 5. **Find the height at $t=4$ minutes:** Substitute $t=4$ into either height equation: $$h_A(4) = 500 - 75 \times 4 = 500 - 300 = 200$$ $$h_B(4) = 100 + 25 \times 4 = 100 + 100 = 200$$ Both birds are 200 feet above the ground after 4 minutes. 6. **Answer the checking questions:** - Bird A is above Bird B when $h_A(t) > h_B(t)$, which is when $t < 4$ minutes. - Bird B is above Bird A when $h_B(t) > h_A(t)$, which is when $t > 4$ minutes. - After 1 minute, find the difference: $$h_A(1) = 500 - 75 = 425$$ $$h_B(1) = 100 + 25 = 125$$ Difference: $$425 - 125 = 300$$ **Final answers:** - Bird A's height: $h_A(t) = 500 - 75t$ - Bird B's height: $h_B(t) = 100 + 25t$ - They are at the same height after 4 minutes. - That height is 200 feet.