MaxWong

Whenever you have $(x + a)^n$ and the exponent is bigger than 2, you want to use binomial theorem to expand it.

Otherwise, you can use sum of square identity (a + b)^2 = a^2 + 2ab + b^2.

$\quad(x - 2)^2 (x + 2)^3 \\= (x^2 - 4x + 4)(\binom{3}0x^3 + \binom{3}1 x^2 \cdot 2+ \binom{3}2 x\cdot 2^2 + \binom{3}32^3) \\= (x^2 - 4x + 4)(x^3 + 6x^2 + 12x + 8)$

Afterwards, you just expand it out like this:

$\quad(x^2 - 4x + 4)(x^3 + 6x^2 + 12x + 8) \\= x^2 (x^3 + 6x^2 + 12x + 8) - 4x (x^3 + 6x^2 + 12x + 8) + 4 (x^3 + 6x^2 + 12x + 8)$

And then expand each clump. It is troublesome, but it will work out nicely.

Common ratio = 10/2 = 5.

First term = 2.

Sum of the first n terms of a G.S. is $\dfrac{a (r^n - 1)}{r - 1}$, where a is the first term and r is the common ratio.

Can you substitute a = 2, r = 5, n = 7 and try to proceed from here?

Let each "equal distance" as L.

The time taken for each "equal distance" is calculated by time = (distance)/(speed) = $\dfrac L{600}, \dfrac L{750}, \cdots, \dfrac L{825}$.

Average speed = (distance)/(time) = $\dfrac{5L}{\dfrac L{600} + \dfrac L{750} + \dfrac L{800} + \dfrac L{650} + \dfrac L{825}} = \dfrac 5{\dfrac1{600} + \dfrac 1{750} + \dfrac1{800} + \dfrac1{650} + \dfrac1{825}}$.

Now you can use a calculator to get the required average speed.

For the last 2 terms, you can just replace z^2 by 1 and calculate its value, so that is done.

For the first 2 terms, you want to rewrite it like this: $z + \dfrac1z = \dfrac{z^2}z + \dfrac1z = \dfrac{z^2 + 1}z$

Rewriting and substituting z^2 = 1 gives $\dfrac{1 + 1}z + 1 + \dfrac11 = 2 + \dfrac2z$. This indicates the answer still depends on z.

Note that there are exactly 2 complex numbers satisfying z^2 = 1. (Hint: Move the 1 to the left-hand side and try to factorize.)

Depending on which one you consider, the required value will be different.

The answer Guest gave is not entirely correct due to this reason.

So we know that a unit square is just a 1 by 1 square, so the area of a unit square is 1.

Since rectangle EFGH has 1/5 of the area of the unit square, the area of EFGH is 1/5.

For simplicity, area will be denoted by [] below. (For example, the sentence above translates to [EFGH] = 1/5.)

$[CGF] = \dfrac12 [BCGF]$ (imagine the rectangle BCGF is a piece of sandwich and you want to cut it in halves.)

Similarly, $[AEH] = \dfrac12 [ADHE]$.

Then, we do some algebra.

$\quad [AFCH]\\ = [CGF] + [EFGH] + [AEH]\\ = [EFGH] + \dfrac12 [BCGF] + \dfrac12 [ADHE]\\ = \dfrac15 + \dfrac12 \left([BCGF] + [ADHE]\right)$

Then notice that [BCGF] + [ADHE] is just the remaining area when [EFGH] is taken away from the unit square, so that is 4/5.

$[AFCH] = \dfrac15 + \dfrac12 \cdot \dfrac45 = \text{(you do the remaining calculation)}$ 

Whenever (x', y') is a point on y = f(x), it is always true that y' = f(x').

Therefore $7 = f(9)$.

Now consider the graph of y/2 = f(x) - 3. What happens when we substitute x = 9 on the right-hand side?

$\dfrac y2 = f(9) - 3 = 7 - 3 = 4$

Then y = 8.

Therefore, when x = 9, the point where y = 8 is the point lying on the graph of y/2 = f(x) - 3.

The required point is (9, 8). You can proceed to calculate the sum of the coordinates of that point.

If f(i) = i + 1, the only divisors of i are 1 and i. That means i is a prime. You can calculate the number of primes in the given range for your answer.

Whenever you see a circle in the form $Ax^2 + Ay^2 + Bx + Cy + D = 0$, complete the squares to get something like $(x - h)^2 + (y - k)^2 = r^2$, then the center will just be (h, k) and the radius will just be r.

So, for this one, you move every term to the left-hand side first.

$x^2 + y^2 + 4x + 16y - 18 = 0$

Now, complete the square.

$(x^2 + 4x + 4) - 4 + (y^2 + 16y + 64) - 64 - 18 = 0\\ (x + 2)^2 - 4 + (y + 8)^2 - 64 - 18 = 0\\ (x + 2)^2 + (y + 8)^2 = 86$

Can you take it from here to identify h and k?

Hint: You can write $x + a$ as $x - (-a)$ at all times.

Simplify the equation, and you will get $x^2 + y^2 = 0$. 

This represents the point (0, 0) only, as no other points (x, y) satisfy the equation.

Now it is very obvious what the y-intercept is.

When you move every term to the left-hand side, you get $x^2 + y^2 - 14x - 8y - 13 \leq 0$.

Complete the squares and you will get $(x - 7)^2 + (y - 4)^2 \leq 78$.

So this is actually just the area enclosed by the circle centered at (7, 4) and with radius $\sqrt{78}$. Can you take it from here?