MaxWong

$0\le \theta < 2\pi\\ \implies0\le \dfrac{2\theta}{3} <\dfrac{4\pi}{3}$

$\cot \left(\dfrac{2\theta}{3}\right)= -\sqrt3\\ \tan\left(\dfrac{2\theta}{3}\right) = -\dfrac{1}{\sqrt3}\\ \dfrac{2\theta}{3} = \dfrac{5\pi}{6}\\ \theta = \dfrac{5\pi}{4}$

1 + rt2 is a root => 1 - rt 2 is also a root

1 + rt3 is a root => 1 - rt 3 is also a root

P(x) = $((x-1)^2 - 2)((x-1)^2 - 3)$

P(x) = $(x-1)^4 - 5(x-1)^2 + 6$

P(x) = $x^4-4x^3+6x^2-4x+1-5x^2+10x-5+6$

P(x) = $x^4-4x^3+x^2+6x+2$

Q: Solve cos x = 0.4 for 0 rad < x < 60 (rad?)

$\cos x = 0.4\\ x = 2n\pi + \arccos \left(0.4\right)\text{ or }x = 2n\pi - \arccos(0.4)$

Just find out how many of these general solutions fit inside the range.

$\text{When }ax^2 + bx + c = 0\text{ has only 1 solution, }b^2 = 4ac$

So 

$\left(b+\dfrac 1 b\right)^2 = 4c\\ b^2 + \dfrac 1 {b^2} - 4c + 2 = 0\\ \left(b^2\right)^2 - (4c - 2)b^2 +1 = 0\\ $

And there is only 1 positive value for b, so there are only 1 possible value for b.

Use that again and you get: 

$\left(-(4c-2)\right)^2 = 4\\ 2c - 1 = 1\\ c = 1$

I am not sure about what the question is talking about, but I guess you meant that. 

$3x + \dfrac{2}{x} - 4 = 2(x+9) - (7-\dfrac 1 x)\\ 3x + \dfrac{2}{x} - 4 = 2x+18-7+\dfrac{1}{x}\\ x + \dfrac{1}{x} = 15$

$V = 150\pi\\ \pi r^2 h = 150\pi\\ \pi \left(\dfrac{5h}{6}\right)^2 h = 150\pi\\ \dfrac{25h^3}{36} = 150\\ h = \sqrt[3]{\dfrac{150\cdot 36}{25}} = 6\\ r = \dfrac{5h}{6} = 5\\ \text{Total surface area} = 2\pi r (r + h) = 2\pi (5)(5+6) = 110\pi$

When a = 1 1/2 = 3/2,

$2a = 3\\ 2a - 3 = 0$

$(5a^2-13a+4)(2a - 3) = (5a^2-13a+4)(0) = 0$

Make use of the identity: $z\cdot z^{*} = |z|^2$.

$(z-5-i)(z-5-i)^* = 25$

Let z = x + y i.

$((x-5)+(y-1)i)((x-5)+(1-y)i) = 25\\ (x-5)^2 + (y-1)^2 = 25$

So the equation |z - 5 - i| = 5 actually represents a circle centered at z = 5 + i and with radius 5.

Expressing the given expression with x and y:

$|z-1+2i|^2 + |z-9-4i|^2\\ =(x-1)^2+(y+2)^2 + (x-9)^2 + (y-4)^2\\ = x^2-2x+1+y^2+4y+4+x^2-18x+81+y^2-8y+16\\ =2x^2+2y^2-20x-4y+102\\ =2(x^2+y^2-10x-2y+102)\\ =2((x-5)^2+(y-1)^2+76)$

And because $(x-5)^2+(y-1)^2 = 25$,

$|z-1+2i|^2 + |z-9-4i|^2\\ =2(25+76)\\ =202$

So the minimum value is 202.

$z^5 = 1 \implies z = e^{\frac{2k\pi i}{5}}, k = 1,2,3,4\\ \dfrac{1}{z} = z^4\\ \dfrac{1}{z^2} = z^3\\ \text{Substitute }z = e^{\frac{2\pi i}5}\\ z+\dfrac1{z}+z^2+\dfrac1{z^2} = z+z^2+z^3+z^4 = \displaystyle \sum_{k=1}^4 e^{\frac{2k\pi i}{5}}$

So the question is actually asking the sum of roots of the equation $z^5-1 = 0$, excluding 1.

Sum of roots of $ax^5+bx^4+cx^3+dx^2+ex+f=0$ is $\dfrac{-b}{a}$. But don't forget to exclude 1.

The required value = $-\dfrac{0}{1} - 1$ = -1.

You can learn from other's solution. Come on!

https://web2.0calc.com/questions/please-help-asap_88