answer:1. **Rewrite the quadratic equation**: The given quadratic equation is ( y = x^2 - 2px + q ). We can complete the square to rewrite it in vertex form: [ y = x^2 - 2px + q = (x - p)^2 + q - p^2 ] This shows that the vertex of the parabola is at ( (p, q - p^2) ). 2. **Condition for no intersection with the x-axis**: For the graph of ( y = x^2 - 2px + q ) to not intersect the x-axis, the discriminant must be negative: [ Delta = (-2p)^2 - 4 cdot 1 cdot q = 4p^2 - 4q < 0 implies p^2 < q ] 3. **Points ( A ) and ( B ) on the graph**: For ( AB ) to be parallel to the x-axis, the y-coordinates of ( A ) and ( B ) must be the same. Let ( A = (p - a, f(p - a)) ) and ( B = (p + a, f(p + a)) ). Then: [ f(p - a) = (p - a)^2 - 2p(p - a) + q = a^2 - p^2 + q ] [ f(p + a) = (p + a)^2 - 2p(p + a) + q = a^2 - p^2 + q ] Thus, ( A = (p - a, a^2 - p^2 + q) ) and ( B = (p + a, a^2 - p^2 + q) ). 4. **Orthogonality condition**: The points ( A ) and ( B ) form vectors from the origin ( O ). For ( angle AOB = frac{pi}{2} ), the vectors ( overrightarrow{OA} ) and ( overrightarrow{OB} ) must be orthogonal. The dot product of these vectors must be zero: [ overrightarrow{OA} = (p - a, a^2 - p^2 + q), quad overrightarrow{OB} = (p + a, a^2 - p^2 + q) ] [ (p - a)(p + a) + (a^2 - p^2 + q)^2 = 0 ] Simplifying the dot product: [ (p^2 - a^2) + (a^2 - p^2 + q)^2 = 0 ] [ (a^2 - p^2 + q)^2 = a^2 - p^2 ] 5. **Solve for ( q )**: Let ( b = a^2 - p^2 ). Then the equation becomes: [ (b + q)^2 = b ] [ b^2 + 2bq + q^2 = b ] [ b^2 + (2q - 1)b + q^2 = 0 ] For there to be a unique solution for ( b ), the discriminant of this quadratic equation must be zero: [ (2q - 1)^2 - 4q^2 = 0 ] [ 4q^2 - 4q + 1 - 4q^2 = 0 ] [ -4q + 1 = 0 ] [ q = frac{1}{4} ] The final answer is ( boxed{frac{1}{4}} ).

answer:We need to ensure all expressions under square roots are nonnegative: 1. sqrt{x} is defined and nonnegative if x geq 0. 2. For sqrt{7 - sqrt{x}} to be defined, we require: 7 - sqrt{x} geq 0 sqrt{x} leq 7 x leq 49 3. For sqrt{3 - sqrt{7 - sqrt{x}}}, we need: 3 - sqrt{7 - sqrt{x}} geq 0 sqrt{7 - sqrt{x}} leq 3 By squaring both sides, 7 - sqrt{x} leq 9 -sqrt{x} leq 2 sqrt{x} geq -2 (which is always true since sqrt{x} geq 0) Combining these inequalities, the most restrictive valid condition from step 2 gives us x leq 49 and x geq 0. Therefore, the domain of g(x) is: boxed{[0, 49]}

answer:To solve this problem, we start by understanding the relationships between the number of fruits in each bucket. Let's denote the number of fruits in bucket C as C, in bucket B as B, and in bucket A as A. Given that bucket C has 9 pieces of fruit, we can write: [C = 9] Since bucket B has 3 more pieces of fruit than bucket C, we can express this as: [B = C + 3] Substituting the value of C into this equation gives us: [B = 9 + 3 = 12] Similarly, bucket A has 4 more pieces of fruit than bucket B, which can be written as: [A = B + 4] Substituting the value of B into this equation gives us: [A = 12 + 4 = 16] To find the total number of pieces of fruit in all three buckets, we add the number of fruits in each bucket: [Total = A + B + C = 16 + 12 + 9] [Total = 37] Therefore, the total number of pieces of fruit in all three buckets is boxed{37}.

answer:Given the root 3 - 2i, the polynomial must also have the complex conjugate as a root because it has real coefficients. Hence, the other root is 3 + 2i. The polynomial can be found by: 1. Writing down (x - (3 - 2i))(x - (3 + 2i)). 2. Simplifying this expression: [(x - 3 + 2i)(x - 3 - 2i).] Using the identity (a + b)(a - b) = a^2 - b^2, we get: [(x - 3)^2 - (2i)^2 = (x^2 - 6x + 9) - 4i^2.] Note that (i^2 = -1) hence (4i^2 = -4), resulting in: [(x^2 - 6x + 9) + 4 = x^2 - 6x + 13.] Thus, the monic quadratic polynomial with real coefficients and roots 3 - 2i and 3 + 2i is (boxed{x^2 - 6x + 13}).