SparklingWater2

77x9/11=63. 

77-63=$\boxed{14}$

Parabola

We can factor N^2+8N+15 into (N+3)(N+5). So if we want N^2+8N+15 to be divisible by 7, we need either n+3 or N+5 divisible by 7. The list of N such that N+3 is divisible by 7 is: 4,11,18,25,...,95, which is (95-4)/7+1=14 numbers. The list of N such that N=5 is divisible by 7 is 2, 9, 16, 23,...,100 which is (100-2)/7+1=15 numbers. 14+15=$\boxed{29}$

We can see that n+2 is divisible by 3, so the lowest possible n that fits this is 2023. We know that 3n+4 is divisible by 5, so n has to end with 2 or 7. The lowest possible value for both statements to hold true is 2032 and we know that 4n+5 is divisible by 9, so the sum of 4n+5's digits have to be divisble by 9 as well, so the sum of 4n's digits have to be 5 less than a multiple of 9. Without the other two statements, the lowest value of this is 2026. You will realize that every single time you add 9 to this number, the property still holds that 4n+5 is divisible by 9. Same goes for the number 2032. So through trial and error we get the lowest number of n is 

$\boxed{2062}$

We know that the three angles of a triangle sum up to 180, so angle Z = 75. We know that X = 60, so the angle bisector must split YXZ into 30 and 30. Now we can see that the angle bisector has split triangle XYZ into two smaller triangles, one being XWZ. Again, we can use our knowledge of the sum of triangle angles, so X + W + Z = 180 = 30 + 75 + W

So angle XWZ = $\boxed{95}$

The first quadratic simplifies to 20x^2 + 30x + 26, and the second one expands to m^2x^2 + 2mnx + n^2. As the constant terms are the only terms that are different between these two, we know that 2mnx = 30x. Simplifying, we get $mn=\boxed{15}$

If we need a binomial, then that means we need only 2 terms. One being $36x^2$, and the other being an integer. This means we have to cancel out the terms with a variable of x. 

Lets let the missing integer be y:

$(3x+1)(12x-y)=36x^2-3xy+12x-y$

We just need to find an integer y such that $-3xy+12x=0$ We can divide both sides by -3 and we see that y has to equal $\boxed4$

If there are 10 cups then the chance of a ping pong ball landing in any cup is$\frac1{10}$

The first ball can go in any cup, but one of the next balls has to land in that same cup, and one has to land in a different cup. So the probability of this is ${\frac1{10}\cdot\frac9{10}=\boxed{\dfrac9{100}}}$

For each value of $x,$ there is a negative value of $x.$ So, the answer to this problem is $\boxed0$

$\frac9{\frac34}=\frac{36}3=\boxed{12}$

We can add the 3 numbers together:

$\frac{64}3+17\frac23+18\frac15=$

$21\frac13+17\frac23+18\frac15=$

$39+18\frac15=\boxed{57\frac15}$