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knowledge drill 2 3 true false activity - RTA

📅 May 2, 2026 👤 scraface

May 02, 2026

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📌 s = (600\sqrt{2})^{1/3}

📌 Thus, the side length \( s \) is \(\boxed{(600\sqrt{2})^{1/3}}\).

📌 An AI model visualizes data clusters as spheres in 3D space. If the radius of a cluster sphere is reduced by 2 units, its volume decreases by \(128\pi\) cubic units. Find the original radius of the sphere.

📌 Let the original radius be \( r \). The volume of a sphere is:

📌 V = \frac{4}{3}\pi r^3

📌 After reducing the radius by 2, the new radius is \( r - 2 \), and the new volume is:

📌 V_{\text{new}} = \frac{4}{3}\pi (r - 2)^3

📌 The volume decrease is:

🔥 Popular Posts

Better: suppose the sensor grid expands in width and height independen...
Assume: the volume is growing due to dwid/dt and dhi/dt, and length is...
VDot = (dw/dt)(h)(l) + w(dh/dt)(l) + w×h(dl/dt)? But dl/dt not defined...
Assume length is constant, so only dw/dt and dh/dt contribute to dV/dt...
Then dV/dt = (dw/dt)(h)(l) + w(dh/dt)(l) = l [ (dw/dt)h + w(dh/dt) ]
But l unknown.
But V = w×h×l = 1,440 m³
And dV/dt = 1,440 × (dw/dt × (1/h) + w × (dh/dt)/h)? No.
Correct: dV/dt = (dw/dt)(h×l) + (dh/dt)(w×l) + (dw/dt)(dh/dt)×l? No —...
dV/dt = ∂V/∂w × dw/dt + ∂V/∂h × dh/dt + ∂V/∂l × dl/dt

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