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Angle APR==Angle PRQ ==x

Angle PRQ ==Angle QPR==x  {Because triangle PQR is an isosceles triangle}

Angle BPQ ==Angle PQR ==100 + x 

From isosceles triangle PQR, we have:

(100 + x) + x + x ==180

3x ==180 - 100 ==80 

x ==80 / 3 ==26 + 2/3 == 26.6667 degrees

Catherine can roll 150 different rolls.

x = 144/5

Please add the image of the grid.

The smallest possible quotient is:

199 / [1 + 9 + 9] ==199 / 19 ==10.47368421 - the smallest quotient in all 3-digit integers.

Use the formula for the "sum of an arithmetic sequence". Look it up in your Math Textbook.

whats the answer lmao

There are: 6^7 =279,936 possible sequences.

Thanks, that's right!

A coin is tossed 12 times. How many different sequences of tosses could there have been, if the number of heads was at least 10

I've already tried to do (12 choose 10)+(12 choose 11)+1, but I'm not sure exactly what to do since it was wrong

Your solution is correct: [12 C 10] + [12 C 11] + [12 C 12] ==79 / 2^12 ==79 / 4,096

Since we're dealing with coin tosses, the number is actually 12*11 + 12 + 1 = 145.

The number of different sets of three points of this 3-by-3 grid that can be the three vertices of an isosceles triangle is 8. The 8 sets of three points are: (1,1), (3,1), (3,3); (1,1), (3,1), (1,3); (1,1), (1,3), (3,3); (1,3), (3,1), (3,3); (1,1), (2,2), (3,1); (1,1), (2,2), (1,3); (2,2), (3,1), (1,3); (2,2), (3,1), (3,3).

By the quadratic formula, the roots of 7x^2 - x - 3 = 0 are (1 + sqrt(79))/14 and (1 - sqrt(79))/14.

Plugging them into the formula for an infinite geometric series, we get (1 + a + a^2 + a^3 + ...)(1 + b + b^2 + b^3 + ...) = 9/4.