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A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet. 12f 3 ft 31 (a) if water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when h is 1.2 feet deep? X ft/min (b) if the water is rising at a rate of 3/8 inch per minute when h - 2.3, determine the rate at which water is being pumped into the trough fo/min
A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet. 12f 3 ft 31 (a) if water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when h is 1.2 feet deep? X ft/min (b) if the water is rising at a rate of 3/8 inch per minute when h - 2.3, determine the rate at which water is being pumped into the trough fo/min
Therefore ,the water level is rising by 0.1388 ft/min when h is 1.2 feet deep and the water is being pumped atrateof 0.8625 cubic feet per min when water is rising at 3/8 inch/minWhat is volume?A three-dimensionalspace'soccupied volume is measured. It is typically quantified using a variety of imperial or US-standard units as well as SI-derived units. The concept oflengthand volume are related.Here,Given: A trough has a 12 foot length and a 3 foot top width. Three-foot-high isosceles triangles make up its ends. 12f 3 ft 31So the first case:using this we find the rate of Volume,So,=> dV/dt = length * height *dh/dt=> dV/dt = 12 * 1.2 *dh/dt=> 2 = 12*1.2* dh/dt=> 2 = 14.4 * dh/dt=>dh/dt = 2/14.4=>dh/dt = 0.1388 ft/minTherefore ,the water level is rising by 0.1388 ft/min when h is 1.2 feet deep .Second case:Since water level is rising by 3/8 inch per min =0.03125 feet/min=> dV/dt = length * height *dh/dt=> dV/dt = 12 * 2.3 *0.03125=>dV/dt = 0.8625 cubic feet per minTherefore , the water is being pumped at rate of 0.8625 cubic feet per min.To know more aboutvolume, visitbrainly.com/question/13338592#SPJ4...