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:
Train A = 128
Train B = 68
Step-by-step explanation:
We can set up a system of equations for this problem
Let A = # of tons of Train A
Let B = # of tons of Train B
A + B = 196
A = B + 60
Now, we plug in A for the first equation, using substitution
(B+60) + B = 196
2B + 60 = 196
Subtract 60 from both sides
2B = 136
Divide both sides by 2
B = 68
Plug in 68 for B in the 2nd equation
A = 68 + 60
A = 128
Checking work: 128 + 68 = 196 :D hope this helped
It’s A) simplify it you’ll get -42 because a negative times a positive is an negative
1+3+5+7+9+11+13+15+17+19 is the answer. Hopefully this helped!! :)
