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Home » Set this equal to $\mathbfc - \mathbfb = \langle 3, 3, 3 \rangle$: - ToelettAPP

Set this equal to $\mathbfc - \mathbfb = \langle 3, 3, 3 \rangle$: - ToelettAPP

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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🔥 Popular Posts

A_1 = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 6^2 = \frac{\...
If each side is increased by 2 cm, the new side length is \(8\) cm. Th...
A_2 = \frac{\sqrt{3}}{4} \times 8^2 = \frac{\sqrt{3}}{4} \times 64 = 1...
A_2 - A_1 = 16\sqrt{3} - 9\sqrt{3} = 7\sqrt{3} \text{ square cm}
Thus, the area increases by \(\boxed{7\sqrt{3}}\) square cm.
A rectangle with dimensions 3 cm by 4 cm is inscribed in a circle. Wha...
The diagonal of the rectangle is the diameter of the circle. Using the...
d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm}
The radius \(r\) of the circle is half of the diameter:
r = \frac{5}{2} = 2.5 \text{ cm}

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