The engine torque y (in foot-pounds) of one model of car is given by y=−3.75x2+23.2x+38.8, where x is the speed (in thousands of revolutions per minute) of the engine. a. To the nearest ten, what is the engine speed that maximizes torque? revolutions per minute To the nearest hundredth, what is the maximum torque? ft-lbs Question 2

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

step-by-step explanation:

i have done this all on a piece of paper but idk how to put it on here cuz it needs to be drawn

8

step-by-step explanation:

to find the amount of gift bags to use all the supplies, find the gcf or greatest common factor. this is the largest number which multiplies evenly into each term.

48 has factors: 1, 2, 3, 4, 6, 8, 12. 16, 24, 48

80 has factors: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80

64 has factors: 1, 2, 4, 8, 8, 16, 32, 64

the highest number factor they all share in common is 8. this means there will be 8 bags with 6 pencils, 10 markers, and 8 stickers in each bag.