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Thank you! I actually figured the question out after I posted it and I got 

z = 1 + i

z = 1 - i

z = -1 - i

z = -1 + i

Quadratic formula: root of ax^2 + bx + c = 0 is x = (-b pm sqrt(b^2 - 4ac))/2a

Minimum area of triangle PBC is 12.

From the definition of convexity, the set of all possible values of f(1) is (-10/3,2/3).

1. The points that work are C, G, I, and A.

2. The points that work are C and G.

The statement is false.  Counter-examples include f(x) = x^2 and f(x) = x^4.

Tell us what are you trying to do? The calculator can do fractions as it is !! Maybe, you just don't know how to use it??

I'm getting a maximum value of 5.

This one's very tough.

This statement is false.  You can look at counter-examples such as f(x) = x^2 and f(x) = x^4.

By Stewart's Theorem, AD^2 = 68.

We can write -4 in exponential notation as 4e^(pi*i), so the equation is z^4 = 4e^(pi*i).

By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4).  Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i.  Then the other roots work out as

4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,

4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and

4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.

I tried your answer and method. That's incorrect.

You can use the finite difference method: https://en.wikipedia.org/wiki/Finite_difference

The first row is 2, 3, 4, 5, 6, -133

The second row is 1, 1, 1, 1, -139

The rest of the table is

0, 0, 0, -140

0, 0, -140

0, -140

-140

Building back up the table gives f(0) = 125.

+ 4200 x^3 y^3 z^4 

lol this is aops test its easy dont worry

As follows

Part A: 20.9

First, note that there are 2+3+1+4+1=11 players. We add their ages. So 19+19+20+20+20+21+22+22+22+22+23 = 230.

Therefore, the mean age is 230/11 = 20.9090, rounded to 20.9.

Part B: 34

When we get one more member, the player count goes to 12. The mean goes to 22. So 22*12 is the sum of all the player's ages, which is 264. We already know 230 for the 11 original players, so the new player must be 264-230 = 34 years old.

You are very welcome!

:P

\(When \sqrt{(-s)^2+(-8)^2}=10\ why\ is\ s=6\ and -6\ and\ not\ 2? \)

Hello Guest!

\((-s)^2+(-8)^2=100\\ -s=\pm \sqrt{100-64}\\ -s=\pm6\)

\(s_1=6\\ s_2=-6\)         therefore!

  !

One line from the end

\(\displaystyle \frac{1}{-i}=\frac{1}{-i}.\frac{i}{i}=\frac{i}{-1}=-i.\)

Yes that is correct.

Line segments ED and E'D' are congruent. The distance between them is 1 unit.

So, we have (ED')² = (E'D')² + (EE')²

Radius        r = (ED') / 2

( I hope it's clear enough.) 

\(27^x=9^{x+1}\)

Hello Guest!

\((3^3)^x=(3^2)^{x+1}\\ 3^{3x}=3^{2(x+1)}\\ 3x=2x+2\\ \color{blue}x=2\)

  !

Let p and q be the two distinct solutions to the equation (x -5)(2x + 9) = x2 -13x + 40.
What is (p + 3)(q + 3)?

Hello Guest!

\(\color{BrickRed}(x -5)(2x + 9) = x^2 -13x + 40\\ 2x^2+9x-10x-45= x^2 -13x + 40\\ x^2+12x-85=0\)

\(x=-6\pm\sqrt{36+85}\\ x=-6\pm 11\)

\(p=5\\ q=-17\\ \)

\( (p + 3)(q + 3)=8\cdot (-14)\)

\( (p + 3)(q + 3)=-112\)

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