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$\1,3,4\, \2\$ - Portal da AcÃºstica

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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🔥 Popular Posts

P'(x) = -\frac{5000}{x^2} - 0.5 = 0 \quad \text{has no solution}
So no minimum — but this contradicts realistic models.
Alternatively, perhaps the function is \( C(x) = 5000 + 120x - 0.5x^2...
Reinterpreting: "cost per unit" is standard, and we must minimize \( P...
Therefore, likely the quadratic coefficient is positive. Assume a typo...
Alternatively, perhaps it's \( C(x) = 5000 + 120x + 0.5x^2 \)? But it...
Given the instructions, we proceed with the **given** function, but re...
Set to zero: no solution. So no critical point — minimum occurs at bou...
Wait — better: perhaps the cost function is \( C(x) = 5000 + 120x - 0....
Given the confusion, let's assume the function is meant to be minimize...

📝 Recent Posts

An equilateral triangle has a side length of 10 cm. If each side is in...
The original area \(A_1\) of an equilateral triangle with side \(s = 1...
A_1 = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 10^2 = 25\sqr...
The new side length is \(12\) cm, so the new area \(A_2\) is:
A_2 = \frac{\sqrt{3}}{4} \times 12^2 = 36\sqrt{3}
\Delta A = A_2 - A_1 = 36\sqrt{3} - 25\sqrt{3} = 11\sqrt{3}
Thus, the area increases by \(\boxed{11\sqrt{3}}\) square centimeters.
Compute \(\cos 60^\circ\) using the unit circle and verify using the p...
On the unit circle, the coordinates of the point corresponding to \(60...
Alternatively, consider an equilateral triangle with side length 2. By...

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