Tottenham10

Because AB=AD, triangle ABD is isosceles, so angle ABD is equal to angle ADB. We are already given angle BAD is 17 degrees. If we call the measure of angle ABD x, then 2x+17=180, so \(x=\frac{163}{2}\).

Feel free to tell me if I did anything wrong! :D

If we combine like terms, we get \(x^2+12x-4\). By Vieta's, the sum of the roots is -12 and the product is -4. If the roots are a and b, then \((a+b)^2=-12^2=144\). If we subtract twice the product, we can find the sum of the squares of the roots. \(144-(-8)=152\).

Feel free to tell me if I did anything wrong! :D

If we use SFFT to factor this expression, we get \((a+4)(b-3)=119\). Because 119 is 7*17, there are only limited cases. We are trying to get the smallest absolute value, meaning a=13 and b=10. Therefore, the minimal possible value of \(|a-b|\) is 3.

Feel free to tell me if I did anything wrong! :D

\((a+b)^2=a^2+b^2+2ab=6+2ab=16\), so ab=5. If we do \((a+b)^3\), we can expand it to \(a^3+a^2b+ab^2+b^3 \) via the binomial theorem. If we focus on the middle terms, we notice we can turn it into \(ab(a+b) \), which is equal to 5*4=20. Therefore, \(a^3+b^3=64-20=44 \).

Feel free to tell me if I did anything wrong! :D

Let the intersection of SQ and PR be point O, the median of SR be A, and the median of XY be B. Then, the ratio of AO to OB is 28:5 (by similar triangles and AA similarity). Next, we know SM=MP, so if C is the median of PQ, then AB to BC is 1:1, so the ratio of AO to OB to BC is 28:5:33. Therefore, the ratio of AO to OC is 28:38 and the length of PQ is 38.

Feel free to tell me if I made a mistake! :D

The points where the graphs intersect are points that lie on both graphs. Therefore, we can set \(x^3+x^2-3x+5=x^3+2x^2 \). Combining like terms, we get \(x^2+3x-5\). By the quadratic formula, the roots of this equation are \(\frac{-3\pm\sqrt29}{2}\). We have to plug these values back into one of the two equations to figure out the y coordinate. I'll let you figure that out (sry I don't have scratch paper rn).

Feel free to tell me if I did anything wrong! :D

First, we look at 1000 mod 21 and 2000 mod 21. 1000 is congruent to 13 mod 21, and 2000 is congruent to 5 mod 21. We are trying to find how many numbers between 1000 and 2000 are congruent to 3 mod 21. The greatest number is 21*95+3=1998 and the least number is 21*48+3=1011. Because it cycles every 21 numbers, there are 95-48+1=\(48\) numbers that satisfy this property

Feel free to tell me if I did anything wrong! :D

If we rationalize the denominator, we get \(-\frac{\sqrt2-\sqrt3}{1}-\frac{\sqrt2+\sqrt3}{1}\). If we combine like terms, we get \(-2\sqrt2\).

Feel free to tell me if I did anything wrong! :D

I just wanted to note that booboo44's solution may be incorrect. First off, there are 6 choose 2 = 15 possible outcomes, not 30. This is because cases will be counted twice. In booboo44's solution, AB and BA are treated as different cases, when they are two equal cases. The favorable cases should also be 12, not 16. There may be more mistakes later on in the solution (but I won't be covering them). If all calculations are done right, the answer should be 8/15. (I created a solution but apparently it's still being moderated)

This is for educational purposes only. I do not want to start a fight with anyone.

Feel free to tell me if I did anything wrong! :D

If we combine like terms, we get ab-9a-5b+__. Then, we can factor the expression as a(b-9)-5b+__. Because -5b+__ need to be factorable by b-9, the blank must be 45. If we plug it back in, we can factor it into (a-5)(b-9). Therefore, the blank is 45.

Feel free to tell me if I did anything wrong! :D