A fuel oil tank is an upright cylinder, buried so that its circular top is 8 feet beneath ground level. The tank has a radius of 6 feet and is 18 feet high, although the current oil level is only 12 feet deep. Calculate the work required to pump all of the oil to the surface. Oil weighs 50\, \hbox{lb/ft}^3.

  1.504×10⁶ ft·lb

Step-by-step explanation:

We understand the top of the oil in the tank is 12 ft below ground level, and the bottom of the tank is 8+18=26 ft below ground level. Then the average depth of the oil is (12+26)/2 = 19 ft below ground level.

The height of the oil in the tank is 26-12=14 ft, so the volume of it is ...

  V = πr²h = π(6 ft)²(14 ft) = 504π ft³ ≈ 1583.36 ft³

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So, the work required to raise that volume of oil to the surface is ...

  (1538.36 ft³)(50 lb/ft³)(19 ft) = 1.504×10⁶ ft·lb

d

step-by-step explanation:

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.