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Home » \frac1000900 > 1 + 9e^-0.2t \Rightarrow \frac109 > 1 + 9e^-0.2t - Nelissen Grade advocaten

\frac1000900 > 1 + 9e^-0.2t \Rightarrow \frac109 > 1 + 9e^-0.2t - Nelissen Grade advocaten

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 ext{proj}_{\mathbf{c}} \mathbf{v} = rac{-3}{6} egin{pmatrix} 1 \ 2 \ -1 \end{pmatrix} = -rac{1}{2} egin{pmatrix} 1 \ 2 \ -1 \end{pmatrix} = egin{pmatrix} -0.5 \ -1 \ 0.5 \end{pmatrix}

📌 Now compute the perpendicular component:

📌 \mathbf{v}_{\perp} = egin{pmatrix} 2 \ -1 \ 3 \end{pmatrix} - egin{pmatrix} -0.5 \ -1 \ 0.5 \end{pmatrix} = egin{pmatrix} 2.5 \ 0 \ 2.5 \end{pmatrix}

📌 The magnitude of the perpendicular component (which is the desired quantity for speed component):

📌 \|\mathbf{v}_{\perp}\| = \sqrt{(2.5)^2 + 0^2 + (2.5)^2} = \sqrt{6.25 + 6.25} = \sqrt{12.5} = rac{5\sqrt{2}}{2}

📌 Alternatively, using formula for perpendicular component magnitude:

📌 \|\mathbf{v}_{\perp}\| = \sqrt{\|\mathbf{v}\|^2 - \left(rac{|\mathbf{v} \cdot \mathbf{c}|}{\|\mathbf{c}\|}

📌 But since we already computed it:

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A_{\text{original}} = \frac{1}{2} \cdot 9 \cdot 12 = 54 \text{ cm}^2
The area of the equilateral triangle is:
A_{\text{new}} = \frac{\sqrt{3}}{4} \cdot 12^2 = \frac{\sqrt{3}}{4} \c...
The increase in area is:
\Delta A = 36\sqrt{3} - 54
\boxed{36\sqrt{3} - 54}
Question: Compute $\tan 75^\circ$ using angle addition identities.
Solution: Using the identity:
\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}
Let $a = 45^\circ$, $b = 30^\circ$, so:

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