Solveit

First, let's rewrite the equation by moving all terms to one side:

2x^2 + 22 = x^2 - 12x + 22

Subtract x^2 and 22 from both sides:

x^2 + 12x = 0

Factor the left side:

x(x + 12) = 0

So, the solutions are x = 0 and x = -12, which are r and s. We can now find r^2 + s^2:

r^2 + s^2 = (0)^2 + (-12)^2 = 0 + 144 = 144

To find the probability of each event, we need to know the number of tiles with the desired point value and divide it by the total number of tiles. The distribution of tiles and point values is missing in your message, but I can provide the standard distribution used in the game:

Point value | Tiles

1 point | E ×12, A ×9, I ×9, O ×8, N ×6, R ×6, T ×6, L ×4, S ×4, U ×4

2 points | D ×4, G ×3

3 points | B ×2, C ×2, M ×2, P ×2

4 points | F ×2, H ×2, V ×2, W ×2, Y ×2

5 points | K ×1

8 points | J ×1, X ×1

10 points | Q ×1, Z ×1

a. Drawing a tile that is worth 1 point:

There are 12 + 9 + 9 + 8 + 6 + 6 + 6 + 4 + 4 + 4 = 68 tiles worth 1 point.

The probability of drawing a 1-point tile is 68/100 = 17/25.

b. Drawing a tile that is worth 10 points:

There are 1 + 1 = 2 tiles worth 10 points (Q and Z).

The probability of drawing a 10-point tile is 2/100 = 1/50.

d. Drawing a tile that is worth at least 4 points:

To find the probability of drawing a tile worth at least 4 points, we can count the number of tiles worth 4 points or more. There are 2 + 2 + 2 + 2 + 2 + 1 + 1 + 1 + 1 = 14 tiles worth 4 points or more (F, H, V, W, Y, K, J, X, Q, Z). The probability of drawing a tile worth at least 4 points is 14/100 = 7/50.

Let's consider the two colors Sam has as color A and color B. We can list out the possible ways he can color the equilateral triangle:

1. All three sides have color A (AAA).
2. All three sides have color B (BBB).
3. Two sides have color A and one side has color B (AAB).

Note that when the triangle has two sides of color A and one side of color B, there is only one distinct coloring, as rotating or reflecting the triangle would still result in the same AAB configuration.

Therefore, Sam can color the equilateral triangle in 3 different ways.

I think the second one doesn't have solutions right? and the first one is every number that fits 2x+1

Okay, I got it. it is -1, therefore \(-(4)^{50}\)

okay so the answer is 4^{50}, I understood the first one.

for the second one, I cannot divide 35 by 2 or is it what I should do?

yep but opening braces goes like this:

\((2-4*i+2)^{50}\)

I have no idea how to proceed from here and I don't know if I opened the braces correctly

So I came till here, but don't know how to continue:

\(|x+yi+4+3i|\leq|5\cdot(x+yi)+10-9i|\\ |(x+4)+i\cdot(y+3)|\leq|(5x+10)+i\cdot(5y-9)|\\ \sqrt{(x+4)^2+(y+3)^2}\leq\sqrt{(5x+10)^2+(5y-9)^2}\\ x^2+8x+16+y^2+6y+9\leq25x^2+100x+100+25y^2-90y+81\\ x^2+8x+y^2+6y+25\leq25x^2+100x+25y^2-90y+181\\ -24x^2-24y^2-92x+96y\leq156\)

it is like picking and then removing from the deck. So, the second one is also not being put back.

I was writing a long good answer but the page got updated. The above are all the divisors of 7 that you need. If you start counting from 1064 every 11th 7 divisible will be the one that is required. Pay attention to that pattern since it will give you a simpler way to solve the question.

1988-1008 = 980

980/7 = 140 

140 - 8 = 132

so it will start from 1st number required meaning +1

132/11 = 12

Answer: 12+1

1008 1015 1022 1029 1036 1043 1050 1057 1064 1071 1078 1085 1092 1099 1106 1113 1120 1127 1134 1141 1148 1155 1162 1169 1176 1183 1190 1197 1204 1211 1218 1225 1232 1239 1246 1253 1260 1267 1274 1281 1288 1295 1302 1309 1316 1323 1330 1337 1344 1351 1358 1365 1372 1379 1386 1393 1400 1407 1414 1421 1428 1435 1442 1449 1456 1463 1470 1477 1484 1491 1498 1505 1512 1519 1526 1533 1540 1547 1554 1561 1568 1575 1582 1589 1596 1603 1610 1617 1624 1631 1638 1645 1652 1659 1666 1673 1680 1687 1694 1701 1708 1715 1722 1729 1736 1743 1750 1757 1764 1771 1778 1785 1792 1799 1806 1813 1820 1827 1834 1841 1848 1855 1862 1869 1876 1883 1890 1897 1904 1911 1918 1925 1932 1939 1946 1953 1960 1967 1974 1981 1988

Sorry, didn't see that

it is 0