aboslutelydestroying

Let remainder to be r

And k to be a non-negative integer.

Set equation: p = 17k + r

The answer would be: $\boxed{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}$

To check: 

r = 0, k = 1

r = 1, k = 6

r = 2, k = 0

r = 3, k = 0

r = 4, k = 5

r = 5, k = 0

r = 6, k = 1

r = 7, k = 0 

r = 8, k = 3

r = 9, k = 2

r = 10, k = 3

r = 11, k = 0

r = 12, k = 1

r = 13, k = 0

r = 14, k = 1

r = 15, k = 4

r = 16, k = 3

We can use Apollonius's Theorem*

*Apollonius's theorem states that in any triangle, the square of the length of a median is equal to the sum of the squares of the lengths of the two sides containing the median, minus one-quarter of the square of the third side.

So: PM^2 = (2PQ^2 + 2PR^2 − QR^2)/4​

Since PQ = 5, PR = 8, QR = 11

PM^2 = (2*25 + 2*64 - 121)/4

PM^2 = 57/4

PM = $\frac{\sqrt{57}}{2}$

So the answer is $\boxed{\frac{\sqrt{57}}{2}}$

For this question, use formula: (a+b)^2 = a^2+b^2+2ab

since a+b=4

a^2+b^2+2ab = 16

a^2+b^2 = 6 + ab

substitute:

6 + ab + 2ab = 16

3ab + 6 =16

3ab = 10

ab= 10/3

a^3 + b^3 = (a+b)(a^2-ab+b^2)

a+b = 4, a^2+b^2 = 6 + 10/3 = 28/3 , and -ab = -10/3

28/3 - 10/3 = 6

=> 4*6 = 24

here's another way to understand :)

(good job, @NotThatSmart)

Let's consider the card labeled as follows:

Aces: $A_1,A_2,A_3,A_4$

Non-Aces: $B_1,B_2$

The total number of ways to choose 2 cards out of 6 for the first pile is $\binom{6}{2}$. After choosing the first pile, we choose 2 cards out of the remaining 4 for the second pile, which is $\binom{4}{2}$. The remaining 2 cards automatically form the third pile, and this can be done in $\binom{2}{2}$ ways.

Total ways is: $\binom{6}{2}\times\binom{4}{2}\times\binom{2}{2}$

But we have to divide 3!, since the order of the piles doesn't matter.

So we get $\frac{\binom{6}{2}\times\binom{4}{2}\times\binom{2}{2}}{6}$ = 15

Next, we count the number of favorable outcomes where each pile contains exactly one Ace. Each pile must have one Ace and one non-Ace. There are 4 Aces ($A_1,A_2,A_3,A_4$) and 2 Non-Aces ($B_1,B_2$).

The assignment of Aces to each pile can be done in 4 ways for first pile 3 ways for second 2 for final simplifying: $4\times3\times 2$ = 24 ways

Every non-ace gets included ensuring simplified piles count as above:

$\boxed{\frac{4}{15}}$

Apply Burnside's Lemma: Number of distinct colorings = 1/4*(∣C_0 degrees​∣ + ∣C_90 degrees​∣ + ∣C_180 degrees​∣ + ∣C_270 degrees​∣)

For C_0 degrees: No squares are moved so it is 2^16

For C_90 degrees: Since there are 4 quadrants, it is 2^4

For C_180 degrees: Each square must match the one diametrically opposite to it. This creates 8 pairs of squares, where each pair must be the same color, so it is 2^8

For C_270 degrees: You do the same thing as C_90 degrees: 2^4

So the answer is 1/4*(65536 + 16 + 256 + 16) = 16448

(I'm not fully sure about this answer, but if it is wrong please tell me) :)

i dont think you are correct -_-'

thanks! 

$7x^2 + x - 5 = 6x^2 - 11x - 2$ is equal to:
$x^2+12x-3$

$(a - 4) + (b - 4)$

is $a+b-8$

sum of roots: $\frac{-b}{a}$

sum: $\frac{-12}{1}$= -12

-12 - 8 = -20