4 1/6 = 4 7/42
2 4/7 = 2 24/42
2 24/42 + 4 7/42 = 6 31/42
6 31/42 = 283/42
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
Kilograms per cubic centimeter
Answer:
yes
Step-by-step explanation:
We are given y = kxz
We also know 32 = k×6×8, so k = 32/48 = 2/3.
We are then asked to solve:
