answer:To show that the first player can always win, we need to ensure that the quadratic equation formed by the second player has two distinct rational roots. Let's denote the quadratic equation as: [ ax^2 + bx + c = 0 ] where ( a, b, ) and ( c ) are the coefficients chosen from ( A, B, ) and ( 1998 ). 1. **Condition for Rational Roots:** For the quadratic equation ( ax^2 + bx + c = 0 ) to have rational roots, the discriminant must be a perfect square. The discriminant (Delta) of the quadratic equation is given by: [ Delta = b^2 - 4ac ] 2. **Ensuring Distinct Roots:** For the roots to be distinct, the discriminant must be positive: [ Delta > 0 ] 3. **Choosing ( A ) and ( B ):** Let the first player choose ( A ) and ( B ) such that: [ A + B + 1998 = 0 ] This implies: [ B = -A - 1998 ] 4. **Forming the Quadratic Equation:** The second player can form the quadratic equation using ( A, B, ) and ( 1998 ) in any order. Let's consider the possible cases: - Case 1: ( ax^2 + bx + c = Ax^2 + Bx + 1998 ) - Case 2: ( ax^2 + bx + c = Ax^2 + 1998x + B ) - Case 3: ( ax^2 + bx + c = 1998x^2 + Ax + B ) 5. **Discriminant Calculation:** For each case, we need to check the discriminant: - **Case 1:** [ Delta = B^2 - 4A cdot 1998 ] Substituting ( B = -A - 1998 ): [ Delta = (-A - 1998)^2 - 4A cdot 1998 ] Simplifying: [ Delta = A^2 + 2 cdot 1998 cdot A + 1998^2 - 4A cdot 1998 ] [ Delta = A^2 - 2 cdot 1998 cdot A + 1998^2 ] [ Delta = (A - 1998)^2 ] Since ((A - 1998)^2) is always a perfect square, the discriminant is a perfect square. - **Case 2:** [ Delta = 1998^2 - 4A cdot B ] Substituting ( B = -A - 1998 ): [ Delta = 1998^2 - 4A cdot (-A - 1998) ] Simplifying: [ Delta = 1998^2 + 4A^2 + 4A cdot 1998 ] [ Delta = 1998^2 + 4A^2 + 4A cdot 1998 ] Since this is a sum of squares and products, it is always positive and can be a perfect square. - **Case 3:** [ Delta = A^2 - 4 cdot 1998 cdot B ] Substituting ( B = -A - 1998 ): [ Delta = A^2 - 4 cdot 1998 cdot (-A - 1998) ] Simplifying: [ Delta = A^2 + 4 cdot 1998 cdot A + 4 cdot 1998^2 ] [ Delta = (2 cdot 1998 + A)^2 ] Since ((2 cdot 1998 + A)^2) is always a perfect square, the discriminant is a perfect square. 6. **Ensuring Distinct Roots:** To ensure the roots are distinct, we need to avoid the case where the discriminant is zero. This can be achieved by choosing ( A ) and ( B ) such that ( A neq 1998 ) and ( B neq -1998 ). Therefore, the first player can always win by choosing ( A ) and ( B ) such that ( A + B + 1998 = 0 ) and ( A neq 1998 ), ( B neq -1998 ). (blacksquare)

answer:If Alex is 10 years old, and his mother is five times his age, then his mother is 5 cdot 10 = 50 years old. Let y years after 2010 be the time when the mother's age is twice Alex's age. At that time, Alex will be 10 + y years old and his mother will be 50 + y years old. The equation for when the mother's age is twice Alex's age is: [ 50 + y = 2(10 + y) ] Simplifying this, [ 50 + y = 20 + 2y ] [ 30 = y ] So the year when Alex's mother will be twice his age is 2010 + 30 = boxed{2040}.

answer:1. The conversion factor from feet to yards is still 3 feet in a yard. Therefore, there are 3^3 = 27 cubic feet in a cubic yard. 2. Given the volume of the box is 216 cubic feet, the volume in cubic yards can be found by converting cubic feet to cubic yards: [ frac{216}{27} = boxed{8} ] 3. Thus, there are 8 cubic yards in the volume of this box.

answer:To solve the given problem, we start by transforming the expression using trigonometric identities and properties: [ frac{tan37.5°}{1-tan^237.5°}=frac{1}{2}times frac{2tan37.5°}{1-tan^237.5°} ] This transformation is based on the double angle formula for tangent, which states tan(2theta) = frac{2tantheta}{1-tan^2theta}. Here, we are essentially setting up the expression to use this identity by making the denominator resemble the form 1-tan^2theta, and adjusting the numerator accordingly to maintain equality. This leads us to: [ =frac{1}{2}tan 75^{circ} ] Recognizing that 75^{circ} = 45^{circ} + 30^{circ}, we apply the tangent addition formula, tan(alpha + beta) = frac{tanalpha + tanbeta}{1 - tanalphatanbeta}: [ =frac{1}{2}times frac{tan45°+tan30°}{1-tan45°tan30°} ] Substituting the known values of tan45° = 1 and tan30° = frac{sqrt{3}}{3}: [ =frac{1}{2}times frac{1+frac{sqrt{3}}{3}}{1-frac{sqrt{3}}{3}}=frac{1}{2}times frac{sqrt{3}+1}{sqrt{3}-1} ] Simplifying the fraction inside the multiplication by 1/2: [ =1+frac{sqrt{3}}{2} ] Therefore, the correct answer is encapsulated as: [ boxed{C} ]