MaxWong

and where's the shaded region? :)

$\dfrac{12\sqrt 2 \operatorname{cis}\left(7\pi/6\right)}{3 \sqrt{\operatorname{cis}\left(\pi/4\right)}} = \dfrac{12\sqrt 2 \operatorname{cis}\left(7\pi/6\right)}{3 \operatorname{cis}(\pi/8)} = 4 \sqrt 2 \operatorname{cis}\left(7\pi/6 - \pi/8\right) = 4\sqrt{2} \operatorname{cis} (25\pi/24) = -4\sqrt2 \operatorname{cis}(\pi/24)$

$\cos \angle YZX = \dfrac{5^2 + 8^2 - 7^2}{2(5)(8)} = \dfrac12$

Therefore $\angle YZX = 60^\circ$.

Area = $\dfrac12 \cdot 5 \cdot 8 \sin 60^\circ = 10 \sqrt 3$

By tangent properties,

IL = IJ --- (1)

Also LO = LO --- (2) (obvious)

Also IO and JO are both radii, so IO = JO --- (3)

By (1), (2), and (3), and using SSS postulate, $\triangle LIO \cong \triangle LJO$.

Therefore $\angle ILO = \angle JLO$

$\implies$ LO bisects ∠ILJ

In spherical coordinates, the $\phi$ represents the angle between (the line segment between the origin and a point) and the x-y plane.

If $\phi = c$, that means the angle $\phi$ is constant. Construct a line L such that the angle between L and the x-y plane is $\phi$, then revolve it around the z-axis. For any line segment L₀ in the resulting figure, the angle between L₀ and x-y plane is $\phi$. Therefore the resulting figure is a cone.

Therefore the answer is (F).

$2x^{2001} + 3x^{2000} + 2x^{1999} + 3x^{1998} + \cdots + 2x + 3 = 0\\ (2x + 3)(x^{2020} + x^{2018} + \cdots + 1) = 0$

Because $x^{2020} + x^{2018} + \cdots + 1 > 0$, the only possibility is $2x + 3 = 0$.

Solving gives $x = -\dfrac32$.
There is only 1 real root.

Assuming $\sin^{-1} x$ and $\tan^{-1} x$ means inverse trigonometric functions,

$\sin\left(\sin^{-1} \dfrac35 + \tan^{-1} 2\right)\\ = \sin\left(\tan^{-1} \dfrac34 + \tan^{-1}2 \right)\\ = \sin \tan^{-1} \dfrac34 \cos \tan^{-1} 2 + \cos \tan^{-1} \dfrac34 \sin \tan^{-1} 2\\ = \dfrac35 \dfrac1{\sqrt5} + \dfrac45\dfrac{2}{\sqrt5}\\ = \dfrac{11}{5\sqrt5}$

$\log_{y^5}\left(x^3\right) = \dfrac{\log x^3}{\log y^5} = \dfrac{3 \log x}{5 \log y} = \dfrac35 \cdot \log_{y} x$

Therefore k = 3/5.

Lateral area = curved surface area = pi*(8/2)*sqrt(4^2 + 5^2) ~= 80.5 ft^2

$m \angle KML = m \stackrel{\mbox{$\frown$}}{KL} \;= 98^\circ$

Because the angle at center is twice the angle at circumference,

$m\angle KJL = \dfrac12 m\angle KML = 49^\circ$