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There are (312​)=220 ways to choose the three white balls on your ticket. For each of these choices, there is one way to choose the red SuperBall. There is also a way to choose the three white balls and the red SuperBall in the same order that they were drawn, so there are also (312​) winning ticket combinations in which your ticket matches at least two of the white balls drawn.

Finally, there are 8 winning ticket combinations in which your ticket matches the red SuperBall, but does not match any of the white balls drawn. Therefore, there are a total of 220+220+8=448 winning ticket combinations.

Since there are a total of (312​)⋅8=840 ways to choose three white balls and one red ball, the probability that your ticket is a winning ticket is 448/840 = 56/105​​.

Let s=2x3+x−3, then x3+s3=3, or 3s3⋅x3−3s=0. Factoring out x3, we have x3⋅(3s2−1)=0.

Since s=2x3+x−3=0, we have x3=0, or x=0​.

By the quadratic formula, the roots of x2−27​x+413​ are [ \frac{7\pm\sqrt{\left(\frac{7}{2}\right)^2-4\cdot1\cdot\frac{13}{4}}}{2}=\frac{7\pm\sqrt{1}}{2}=\frac{7}{2}\pm\frac{\sqrt{1}}{2} ]

Therefore, the values of x at which the graphs of f and g intersect are x=2, x=27+1​​ and x=27−1​​.

We can plug these x values back into f(x) to get that f(2)=21, f(27+1​​)=825+1​​ and f(27−1​​)=825−1​​, so the points of intersection are [ \left(2,21\right),\left(\frac{7+\sqrt1}{2},\frac{25+\sqrt{1}}{8}\right),\left(\frac{7-\sqrt1}{2},\frac{25-\sqrt{1}}{8}\right). ]

Since the x-intercept of the line is (10,0), we know that the equation of the line is of the form y=mx+n for some value of m and n. Since the y-intercept of the line is (0,−2), we have that −2=m⋅0+n. Hence, n=−2. Therefore, the equation of the line is y=mx−2.

We know that the point (a,a2) lies on the line, so a2=ma−2. We also know that the point (b,b2) lies on the line, so b2=mb−2. Subtracting these two equations, we get b2−a2=m(b−a). Since a 0, so m=b−ab2−a2​.

Since the line passes through the point (10,0), we have that 0=m(10)−2. Substituting the expression for m into this equation, we get 0=b−ab2−a2​(10)−2. This simplifies to b2−a2=5a−5b. We also have the equation a2=ma−2. Substituting m=b−ab2−a2​, we get a2=b−ab2−a2​(a)−2. This simplifies to (b−a)a2−(b2−a2)a+2(b−a)=0. This factors as (a−2)(b−a)a2=0.

Since a

Substituting a=0 into the equation b2−a2=5a−5b, we get b2=5b. This factors as b2−5b=0, which factors as b(b−5)=0. Therefore, b=0 or b=5. Since a

Therefore, (a,b)=(0,5)​.

The distance from the center of the circle to the line is equal to the radius of the circle. To find the radius, we need to find the perpendicular distance from the origin to the line.

The equation of the line in slope-intercept form is y=21​x+215​. The slope of the line is 21​, and its negative reciprocal is −2. Therefore, the equation of the perpendicular line that passes through the origin is y=−2x. Setting the two equations equal to each other, we get −2x=21​x+215​. Solving for x, we get that x=−5.

Substituting this value of x back into the equation of the line, we get that y=5, so the origin is (0,5). Therefore, the radius of the circle is 5, so the area of the circle is π⋅52=25π​ square units.

Case 1: The number has an even number of digits.

Since the sum of the digits is 12, this is impossible.

Case 2: The number has an odd number of digits, and the units digit is 1.

In this case, the first digit must be 3, and the digits in between must add to 9. There are (25​)=10 ways to choose the two numbers that add to 9, and 22=4 ways to order them. Thus the number of possibilities in this case is 10⋅4=40.

Case 3: The number has an odd number of digits, and the units digit is 3.

In this case, the first digit must be 1, and the digits in between must add to 3. There are (13​)=3 ways to choose the number that adds to 3, and 2 ways to place it. Thus the number of possibilities in this case is 3⋅2=6.

The total number of possibilities is 40+6=46​.