MaxWong

You can multiply out just like "i" was a variable instead of sqrt(-1), and then use i^2 = -1 occasionally.

$\quad Q\times E \times D\\ = (11- 5i)(11i)(2i + 3)\\ = (121i - 55i^2)(2i + 3)\\ = (55 + 121i)(2i + 3)$

Can you expand the rest?

First, to find $f^{-1}(x)$, we replace $x$ with $f^{-1}(x)$.

$f(f^{-1}(x)) = 5f^{-1}(x) - 12\\ x = 5f^{-1}(x) - 12\\ f^{-1}(x) = \dfrac{x + 12}5$

Also, what is f(x - 1)? We need to find it as well! Whenever we want to find $f(\triangle)$, we replace x by $(\triangle)$. In this case, replace x with (x - 1).

$f(x - 1) = 5(x - 1) - 12 = 5x - 17$

Then, the equation is just $\dfrac{x + 12}5 = 5x - 17$. I believe you can solve the linear equation on your own.

The absolute value of the amplitude is the coefficient of sin or cos, so either $f(x) = 2 \cos(\boxed{\phantom{ aaaa }}x)+\boxed{\phantom{ aaaa }}$ or $f(x) = -2 \cos(\boxed{\phantom{ aaaa }}x)+\boxed{\phantom{ aaaa }}$. Since it is the reflection of its parent function g(x) = cos(x) over x-axis, we determine that it must be $f(x) = -2 \cos(\boxed{\phantom{ aaaa }}x)+\boxed{\phantom{ aaaa }}$

Vertical shift is the constant term, but you need to consider the direction of the shift. Shifting downwards means a negative constant term, so $f(x) = -2 \cos(\boxed{\phantom{ aaaa }}x)-11$.

We also know that the period of $h(x) = a \cos (bx + c) + d$ is $\dfrac{2\pi}b$. You can solve the equation $\dfrac{2\pi}b = \dfrac{6\pi}7$ to get the remaining blank.

You can choose two non-math books out and place them at the ends first, to ensure math books never will be at the ends.

Number of ways to choose two non-math books = $\displaystyle\binom{6}{2}$.

But you have two ways to choose which book to place at the bottom, so you need to multiply 2.

Then, the remaining 4 books can be rearranged in any order in the middle, so multiply by another $4!$, you get the total number of ways as $720$.

Note that $a \star b$ really looks like $(a + b)^2$, but it is not actually that, so we work around that by doing this:

$a\star b = a^2 + b^2 = (a^2 \color{red} + 2ab\color{black} + b^2) \color{red} - 2ab\color{black} = (a + b)^2 - 2ab$

Now, use a + b = 13 and ab = 25, you will get $a \star b = 119$. Please tell me if you encounter any difficulty in the calculation.

You can just expand it like i was a variable instead of a special number, and use the fact i^2 = -1 occasionally to reduce the power.

$\quad(4 - 5i)(5 + 5i)(2 - 7i)\\ =(20 - 25i + 20i - 25i^2 )(2-7i)\\ = (20 - 5i -25i^2)(2 - 7i)\\ = (45 - 5i)(2 - 7i)\\ $

Can you continue from here, doing what I just did?

Since 13 is a prime, every integer has its inverse mod 13.

We can rewrite this system as 

$\begin{cases} 2a^{-1} + b^{-1} + c^{-1} \equiv 6 \pmod{13}\\ a^{-1} + 2b^{-1} + c^{-1} \equiv 8\pmod{13}\\ a^{-1} + b^{-1} + 2c^{-1} \equiv0 \pmod{13}\\ \end{cases}$

Simplifying gives 

$\begin{cases} a^{-1} + 2b^{-1} + c^{-1} \equiv 8 \pmod{13}\\ -b^{-1} + c^{-1} \equiv 5\pmod{13}\\ -4b^{-1} \equiv8 \pmod{13}\\ \end{cases}$

Then $b^{-1} \equiv 11\pmod{13}$, $c^{-1} \equiv 3 \pmod{13}$, and $a^{-1} \equiv 9\pmod{13}$.

Inverting each of them gives $a \equiv 3\pmod{13}$, $b\equiv 6\pmod{13}$, and $c\equiv9\pmod{13}$.

Then $a + b + c \equiv 5\pmod{13}$. The remainder is 5.

We calculate the area by Heron's formula,

s = (8 + 12 + 14)/2 = 17

Area = sqrt(s(s - 8)(s - 12)(s - 14)) = 3 sqrt(255)

Altitude length corresponding to a side x is 2(area)/x.

Use the shortest side lengths as x, and you will get the longest altitude in result. You can substitute area = 3 sqrt(255) and x = 8, x = 12 to get the length of the two longest altitudes.

Denote M as the midpoint of AB, N as the midpoint of CD.

The answer would be sqrt(PM^2 + PN^2) by Pythagorean theorem.

Note that PM = (AB)/2 - PA = 3.

Let x = PD. Then by power of points, 

$x(13 - x) = 4(10)\\ x^2 - 13x + 40 = 0\\ x = 8\text{ or }x = 5$

Then PC = 8, PD = 5 or PC = 5, PD = 8.

In any case, PN = 13/2 - 5 = 3/2.

Distance from P to the center of the circle = $\sqrt{3^2 + \left(\dfrac32\right)^2} = \dfrac{3\sqrt5}2$