answer:1. **Symmetry Condition**: Given that ( p(x) = p(-x) ) for all ( x ), ( p(x) ) must be an even function. Therefore, ( p(x) ) can be expressed as: [ p(x) = ax^4 + bx^2 + c ] where ( a ), ( b ), and ( c ) are constants. 2. **Non-negativity Condition**: Given that ( p(x) ge 0 ) for all ( x ), the polynomial must be non-negative for all real numbers ( x ). 3. **Value at Zero**: Given that ( p(0) = 1 ), substituting ( x = 0 ) into the polynomial gives: [ p(0) = c = 1 ] Therefore, the polynomial can be rewritten as: [ p(x) = ax^4 + bx^2 + 1 ] 4. **Local Minima Condition**: Given that ( p(x) ) has exactly two local minimum points ( x_1 ) and ( x_2 ) such that ( |x_1 - x_2| = 2 ), we need to find the critical points of ( p(x) ). The critical points are found by setting the first derivative ( p'(x) ) to zero: [ p'(x) = 4ax^3 + 2bx = 0 ] Factoring out ( x ) gives: [ x(4ax^2 + 2b) = 0 ] This equation has solutions: [ x = 0 quad text{or} quad 4ax^2 + 2b = 0 ] Solving ( 4ax^2 + 2b = 0 ) for ( x ) gives: [ 4ax^2 = -2b quad Rightarrow quad x^2 = -frac{b}{2a} ] For ( x ) to be real, ( -frac{b}{2a} ) must be non-negative, implying ( b le 0 ). 5. **Distance Between Minima**: Given ( |x_1 - x_2| = 2 ), and knowing that ( x_1 ) and ( x_2 ) are symmetric about the origin (since ( p(x) ) is even), we can assume ( x_1 = 1 ) and ( x_2 = -1 ). Therefore: [ 1^2 = -frac{b}{2a} quad Rightarrow quad 1 = -frac{b}{2a} quad Rightarrow quad b = -2a ] 6. **Substitute ( b ) into ( p(x) )**: Substituting ( b = -2a ) into the polynomial gives: [ p(x) = ax^4 - 2ax^2 + 1 ] 7. **Non-negativity Check**: To ensure ( p(x) ge 0 ) for all ( x ), we rewrite ( p(x) ) as: [ p(x) = a(x^4 - 2x^2) + 1 = a(x^2 - 1)^2 + 1 - a ] For ( p(x) ge 0 ), the minimum value of ( a(x^2 - 1)^2 ) is 0 (when ( x = pm 1 )), so: [ 1 - a ge 0 quad Rightarrow quad a le 1 ] Since ( a > 0 ) (to ensure the polynomial is of degree 4 and non-negative), we have: [ 0 < a le 1 ] Conclusion: The fourth-degree polynomial ( p(x) ) that satisfies all the given conditions is: [ p(x) = a(x^2 - 1)^2 + 1 - a ] where ( 0 < a le 1 ). The final answer is ( boxed{ p(x) = a(x^2 - 1)^2 + 1 - a } ) where ( 0 < a le 1 ).

answer:From the asymptotes of the hyperbola C: y = pm frac{b}{a}x, We can derive the equation of line l as: y = frac{b}{a}x + 3b, or bx - ay + 3ab = 0, Given that the distance from any point on the right branch of hyperbola C to line l is always greater than b, We can infer that the distance between line l and bx - ay = 0 is always greater than or equal to b, This implies frac{3ab}{sqrt{a^2+b^2}} geqslant b, Simplifying, we get 8a^2 geqslant b^2, Since c^2 = a^2 + b^2, we have 8a^2 geqslant c^2 - a^2, This simplifies to c^2 leqslant 9a^2, hence c leqslant 3a, Thus, the eccentricity e = frac{c}{a} leqslant 3. Therefore, the maximum value of the eccentricity is boxed{3}. To solve this problem, we first find the equation of line l, then use the condition that the distance from any point on the right branch of hyperbola C to line l is always greater than b. We then apply the formula for the distance between parallel lines and establish an inequality to find the maximum value of the eccentricity of hyperbola C. This is a moderately difficult problem that requires a good understanding of hyperbola properties, the formula for point-to-line distance, and algebraic manipulation skills.

answer:Let's denote the amount of darker green paint needed as x gallons. The total amount of yellow paint in the light green paint is 20% of 5 gallons, which is 0.20 * 5 = 1 gallon. The total amount of yellow paint in the darker green paint is 40% of x gallons, which is 0.40 * x = 0.4x gallons. When we mix the two paints, we want to end up with a mixture that is 25% yellow paint. The total amount of paint after mixing will be 5 gallons + x gallons = (5 + x) gallons. The total amount of yellow paint in the mixture will be the sum of the yellow paint from the light green paint and the yellow paint from the darker green paint, which is 1 gallon + 0.4x gallons. The equation representing the final mixture being 25% yellow paint is: 1 + 0.4x = 0.25 * (5 + x) Now, let's solve for x: 1 + 0.4x = 0.25 * 5 + 0.25x 1 + 0.4x = 1.25 + 0.25x 0.4x - 0.25x = 1.25 - 1 0.15x = 0.25 x = 0.25 / 0.15 x = 5/3 x = 1.6667 gallons Therefore, you would need approximately 1.6667 gallons of the darker green paint to create a green paint that is boxed{25%} yellow paint.

answer:First, let's calculate the total number of pregnancies across all 6 kettles: 6 kettles * 15 pregnancies per kettle = 90 pregnancies Now, let's calculate the total number of babies expected from these pregnancies before any losses: 90 pregnancies * 4 babies per pregnancy = 360 babies However, since approximately 25% are lost, we need to calculate the number of babies that survive: 75% of 360 babies = 0.75 * 360 = 270 babies Therefore, the ornithologists can expect approximately boxed{270} hawk babies this season.