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 = \(\sqrt{3^2 + \left(\dfrac32\right)^2} = \dfrac{3\sqrt5}2\)