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Home » Solve: \(x = 1.5\). - Portal da AcÃºstica

Solve: \(x = 1.5\). - Portal da AcÃºstica

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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🔥 Popular Posts

d = 4
Thus, the shortest distance from the center to the path is $ \boxed{4}...
Question: A robotics engineer is designing a triangular linkage mechan...
Let the triangle have side lengths $ a = 7 $, $ b = 10 $, $ c = 13 $....
s = \frac{a + b + c}{2} = \frac{7 + 10 + 13}{2} = \frac{30}{2} = 15
Now, use Heron’s formula to find the area $ A $:
A = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{15(15 - 7)(15 - 10)(15 - 13)...
A = \sqrt{15 \cdot 8 \cdot 10} = \sqrt{1200} = \sqrt{400 \cdot 3} = 20...
r = \frac{A}{s} = \frac{20\sqrt{3}}{15} = \frac{4\sqrt{3}}{3}
Thus, the radius of the inscribed circle is $ \boxed{\dfrac{4\sqrt{3}}...

📝 Recent Posts

s = \sqrt{64} = 8 \text{ units}
Increasing each side by \(50\%\) gives a new side length:
s_{\text{new}} = 8 + 0.5 \times 8 = 1.5 \times 8 = 12 \text{ units}
The new area is:
A_{\text{new}} = (12)^2 = 144 \text{ square units}
Alternatively, since area scales by the square of the side length fact...
A_{\text{new}} = 64 \times 2.25 = 144
The new area of the square is \(\boxed{144}\) square units.
Compute \(\cos 300^\circ\).
The angle \(300^\circ\) lies in the fourth quadrant, where cosine is p...

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