SuperBrain1 From: US Posts: 10 Votes: 20 |
The volume of a cube is decreasing at the rate of 10m^3/hr. How fast is the total surface area decreasing when the surface area is 54m^2? |

DedicatedRoger From: US Posts: 33 Votes: 33 |
--A = 6s^2 V = s^3 A = 54 54 = 6s^2 9 = s^2 3 = s V = s^3 dV/dt = 3s^2 * ds/st dV/dt = -10 -10 = 3 * 3^2 * ds/dt -10 = 27 * ds/dt -10/27 = ds/dt A = 6s^2 dA/dt = 12s * ds/st dA/dt = 12 * 3 * (-10/27) dA/dt = -120 / 9 dA/dt = -40/3 |

Anonymous From: - Posts: - Votes: - |
Sorry i put wrong answer please donot consider it |

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