A cable that weighs 8 lb/ft is used to lift 650 lb of coal up a mine shaft 600 ft deep. Find the work done. Show how to approximate the required work by a Riemann sum. (Let x be the distance in feet below the top of the shaft. Enter xi* as xi.)

1830000 ft*lb

Step-by-step explaination:

Work done = fd

f=force:

d = distance

f = 650 + 8x lbs, x is current amount of cable in shaft

dW = fdx = (650+8x)dx

x ranges from 0 to 600 then the total work is:

W = integral(x=0 to x=600)((650+8x)dx)

W = (650x+4x^2)[x=0, x=600]

W = (650*600 + 4*(600)^2) = 1830000 ft*lb

W = 1830000 ft*lb

Riemann sum approximation:

W = sum(i=1 to i=n)(650+8xi*)(xi - x(i-1))

x0 is 0, xn is 600 (or x0 is 600, xn is 0)

assuming the partition is a constant interval deltax sum can be written as:

W = summation(i=1 to i=n)(650+8xi*)deltax

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!

cheater

step-by-step explanation: