%3Cp%3EKindly%20solve%26nbsp%3Bthe%26nbsp%3Bboth%26nbsp%3Bof%26nbsp%3Bthe%26nbsp
ID: 2844326 • Letter: #
Question
%3Cp%3EKindly%20solve%26nbsp%3Bthe%26nbsp%3Bboth%26nbsp%3Bof%26nbsp%3Bthe%26nbsp%3Bparts%26nbsp%3Bof%26nbsp%3Bthe%26nbsp%3Bword%26nbsp%3Bproblem%26nbsp%3Brelated%26nbsp%3Bto%26nbsp%3Bdifferential%26nbsp%3Bequation%26nbsp%3Bgiven%26nbsp%3Bin%26nbsp%3Bthe%26nbsp%3Bpicture.%26nbsp%3BThank%26nbsp%3Byou!%3C%2Fp%3E%3Cp%3E%3Cimg%26nbsp%3Bclass%3D%22user-upload%22%26nbsp%3Bsrc%3D%22http%3A%2F%2Fmedia.cheggcdn.com%2Fmedia%252Fad6%252Fad6a4a34-263c-43f7-aaf3-125bc7cdc58b%252FphpwV42ep.png%22%26nbsp%3Bdata-mce-src%3D%22http%3A%2F%2Fmedia.cheggcdn.com%2Fmedia%252Fad6%252Fad6a4a34-263c-43f7-aaf3-125bc7cdc58b%252FphpwV42ep.png%22%26nbsp%3Bwidth%3D%22960%22%26nbsp%3Bheight%3D%22720%22%3E%3C%2Fp%3EExplanation / Answer
Y = height of water a time t
= 12 -k*t
given at t = 1 hour , y = 6 feet
hence k = 12-6 = 6
1) when y = 6 , then X = [6]^3/4 = 3.83
area =pi*x^2 = 3.14*(3.83)^2 = 46.148
2) time required to emepty = 12/k = 2 hour
hence tank will emepty at 2 pm
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