Calculus (Definite Integrals - Force) Please Help
A cylindrical oil storage tank 12 feet in diameter and 17feet long is lying on its side. Suppose the tank is half full of oil weighing 85 lb per cubic foot. What's the total force on one endof the tank?
CALCULUS
A cylindrical oil storage tank 12 feet in diameter and 17feet long is lying on its side. Suppose the tank is half full of oil weighing 85 lb per cubic foot. What's the total force on one endof the tank?
Calculus (Definite Integrals - Work)
Recall that work is defined to be force times distance, and that the weight (force) of a liquid is equal to its volume times its density. A fish tank has a rectangular base of width 2 feet and length 6 feet and sides of height 5 feet. If the tank is filled with water weighing ...
Calculus (Definite Integrals - Arclength)
Using the trapezoid rule with n = 8 to approximate the arc length of the graph of y = 2x^3 - 2x + 1 from A(0,1) to B(2,13) you get (to three decimal places): A.) 6.900 B.) 13.896 C.) 14.093 D.) 13.688 E.) 13.697
Calculus
Using the trapezoid rule with n = 8 to approximate the arc length of the graph of y = 2x^3 - 2x + 1 from A(0,1) to B(2,13) you get (to three decimal places): A.) 6.900 B.) 13.896 C.) 14.093 D.) 13.688 E.) 13.697
CALCULUS
In Seattle on September 30, the temperature hours after midnight was given by the function y=60 + 12sin((pi/x)(x-11)) What was the average temperature over the period from 8 A.M. until 10 P.M.?
Calculus (Definite Integrals)
How many definite integrals would be required to represent the area of the region enclosed by the curves y=(cos^2(x))(sin(x)) and y=0.03x^2, assuming you could not use the absolute value function? a.) 1 b.) 2 c.) 3 d.) 4 e.) 5
Calculus (Area Between Curves)
Find the area of the region IN THE FIRST QUADRANT (upper right quadrant) bounded by the curves y=sin(x)cos(x)^2, y=2xcos(x^2) and y=4-4x. You get: a.)1.8467 b.) 0.16165 c.) 0.36974 d.) 1.7281 e.) 0.37859
Calculus (Area Between Curves)
Thank you, I was just really unsure of my answer!
Calculus (Area Between Curves)
Find the area of the region IN THE FIRST QUADRANT (upper right quadrant) bounded by the curves y=sin(x)cos(x)^2, y=2xcos(x^2) and y=4-4x. You get: a.)1.8467 b.) 0.16165 c.) 0.36974 d.) 1.7281 e.) 0.37859 Based on my calculations, I would say that the answer is e.) 0.37859. I a...
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