the distance between players is the hypotenuse of.pdf
the distance between players is the hypotenuse of a right triangle. Legs from the right angle are the runners distance from second. runner heading to second distance = 90 - 18t runner running from second = 20 t Distance between runners = D D^2 = (20t)^2 + (90 - 18t)^2 D^2 = 724t^2 - 3240t + 8100 D = sqrt(724t^2 - 3240t + 8100) <== equation to differentiate dD/dt = (362t - 810) / sqrt(181t^2 - 810t + 2025) D \' (1) = (362 - 810) / sqrt(181 - 810 + 2025) = -11.99 ft/sec ( approx) negative indicates distance is decreasing. Solution the distance between players is the hypotenuse of a right triangle. Legs from the right angle are the runners distance from second. runner heading to second distance = 90 - 18t runner running from second = 20 t Distance between runners = D D^2 = (20t)^2 + (90 - 18t)^2 D^2 = 724t^2 - 3240t + 8100 D = sqrt(724t^2 - 3240t + 8100) <== equation to differentiate dD/dt = (362t - 810) / sqrt(181t^2 - 810t + 2025) D \' (1) = (362 - 810) / sqrt(181 - 810 + 2025) = -11.99 ft/sec ( approx) negative indicates distance is decreasing..
the distance between players is the hypotenuse of.pdf
the distance between players is the hypotenuse of a right triangle. Legs from the right angle are the runners distance from second. runner heading to second distance = 90 - 18t runner running from second = 20 t Distance between runners = D D^2 = (20t)^2 + (90 - 18t)^2 D^2 = 724t^2 - 3240t + 8100 D = sqrt(724t^2 - 3240t + 8100) <== equation to differentiate dD/dt = (362t - 810) / sqrt(181t^2 - 810t + 2025) D \' (1) = (362 - 810) / sqrt(181 - 810 + 2025) = -11.99 ft/sec ( approx) negative indicates distance is decreasing. Solution the distance between players is the hypotenuse of a right triangle. Legs from the right angle are the runners distance from second. runner heading to second distance = 90 - 18t runner running from second = 20 t Distance between runners = D D^2 = (20t)^2 + (90 - 18t)^2 D^2 = 724t^2 - 3240t + 8100 D = sqrt(724t^2 - 3240t + 8100) <== equation to differentiate dD/dt = (362t - 810) / sqrt(181t^2 - 810t + 2025) D \' (1) = (362 - 810) / sqrt(181 - 810 + 2025) = -11.99 ft/sec ( approx) negative indicates distance is decreasing..
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the distance between players is the hypotenuse of.pdf
- 1. the distance between players is the hypotenuse of a right triangle. Legs from the right angle are the runners distance from second. runner heading to second distance = 90 - 18t runner running from second = 20 t Distance between runners = D D^2 = (20t)^2 + (90 - 18t)^2 D^2 = 724t^2 - 3240t + 8100 D = sqrt(724t^2 - 3240t + 8100) <== equation to differentiate dD/dt = (362t - 810) / sqrt(181t^2 - 810t + 2025) D ' (1) = (362 - 810) / sqrt(181 - 810 + 2025) = -11.99 ft/sec ( approx) negative indicates distance is decreasing. Solution the distance between players is the hypotenuse of a right triangle. Legs from the right angle are the runners distance from second. runner heading to second distance = 90 - 18t runner running from second = 20 t Distance between runners = D D^2 = (20t)^2 + (90 - 18t)^2 D^2 = 724t^2 - 3240t + 8100 D = sqrt(724t^2 - 3240t + 8100) <== equation to differentiate dD/dt = (362t - 810) / sqrt(181t^2 - 810t + 2025) D ' (1) = (362 - 810) / sqrt(181 - 810 + 2025) = -11.99 ft/sec ( approx) negative indicates distance is decreasing.