Answer:
Step-by-step explanation:
If the engine torque y (in foot-pounds) of one model of car is given by y=−3.75x^2+23.2x+38.8
The engine speed is at maximum if dy/dx = 0
dy/dx = -2(3.75)x+23.2
dy/dx = -7.5x + 23.2
since dy/dx = 0
0 = -7.5x + 23.2
7.5x = 23.2
x = 23.2/7.5
x = 3.093
Hence the maximum torque is 3.09 rev/min
Answer:
this is simplify 1/16 x4 y4
If it is asking if that equation is the quadratic formula, then the answer is false. The reason why is that the first 'b' should be negative
The quadratic formula is
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