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:
10.9
Step-by-step explanation:
adding there squres you get 118.1 taking the squareroot of that you get 10.9
The equation in slope intercept form would be: y= 1/26x - 2
Suppose m∠1 = x degree
m∠2 = 17 x degree
As angle 1 & 2 are supplementary angles so
m∠1 +m∠2 =180 degree...... eq 1
Substituting the values of angle 1 & 2 in eq 1, we get
x +17x =180
18x=180
x= 180/18 =10 degree
17 x= 170 degree
m∠1 = 10 degree m∠2 = 170 degree.
Answer:
A: 22
Step-by-step explanation:
2The interquartile range begins at 45 and ends at 67. All you need to do is subtract 45 drom 67, and you get 22.
