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Home » \(x = -\frac802 \times -2 = 20\). - Aurero

\(x = -\frac802 \times -2 = 20\). - Aurero

📅 March 13, 2026 👤 scraface

Mar 13, 2026

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📌 A spherical balloon is being inflated such that its radius increases at a constant rate of 0.5 cm/s. At what rate is the volume increasing when the radius is 10 cm?

📌 Volume of a sphere:

📌 Differentiate with respect to time \( t \):

📌 \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

📌 Plug in \( r = 10 \) cm and \( \frac{dr}{dt} = 0.5 \) cm/s:

📌 \frac{dV}{dt} = 4\pi (10)^2 (0.5) = 4\pi \times 100 \times 0.5 = 200\pi \text{ cm}^3/\text{s}

📌 The position of a particle moving along a line is given by \( s(t) = t^3 - 6t^2 + 9t + 2 \), where \( s \) is in meters and \( t \) in seconds. Find the time(s) when the particle is at rest.

🔥 Popular Posts

Increased by 80%: 12 × 1.8 = <<12*1.8=21.6>>21.6 ms
Processing: 60% of 48 = 0.6 × 48 = <<0.6*48=28.8>>28.8 ms
Rendering: 15% of 48 = 0.15 × 48 = <<0.15*48=7.2>>7.2 ms
New total: 28.8 + 21.6 + 7.2 = <<28.8+21.6+7.2=57.6>>57.6 ms
#### 57.6**Question:** Compute \(\cos 45^\circ + \cos 225^\circ\).
First, we calculate \(\cos 45^\circ\) and \(\cos 225^\circ\) separatel...
\cos 45^\circ = \frac{\sqrt{2}}{2}
For \(\cos 225^\circ\), since \(225^\circ = 180^\circ + 45^\circ\), we...
\cos 225^\circ = \cos(180^\circ + 45^\circ) = -\cos 45^\circ = -\frac{...
Now, adding the two values:

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