answer:Let's denote the speed of the stream as ( v ) km/hr. When the boat is going downstream, the speed of the boat and the speed of the stream add up. So, the effective speed of the boat downstream is ( (24 + v) ) km/hr. We are given that the boat takes 5 hours to travel 140 km downstream. We can use the formula for speed, which is distance divided by time, to find the effective speed of the boat downstream: [ text{Speed} = frac{text{Distance}}{text{Time}} ] Plugging in the given values: [ 24 + v = frac{140}{5} ] [ 24 + v = 28 ] Now, we can solve for ( v ): [ v = 28 - 24 ] [ v = 4 ] km/hr So, the speed of the stream is boxed{4} km/hr.

answer:(I) Since the domain of the function f(x)=sqrt{2|x-3|-|x|-m} is mathbb{R}, we must have 2|x-3|-|x|-mgeq 0 for all x. This translates to requiring m leq min(2|x-3|-|x|). To find this minimum value, consider y = 2|x-3|-|x|: - For x > 3, we have y = 2x - 6 - x = x - 6 which implies y geq -3; - For x < 0, we get y = 6 - 2x + x = 6 - x, giving y geq 6; - For 0 leq x leq 3, we have y = 6 - 2x - x = 6 - 3x, meaning -3 leq y leq 6. Thus, we see that y is minimized for x=3, at which point y = -3. Consequently, the range of m is (-infty, -3]. (II) Since a^2+b^2+c^2=t^2, given that the maximum value of m is t, we have a^2+b^2+c^2=9. Let u=a^2+1, v=b^2+2, and w=c^2+3, which yields u+v+w=15. Then, frac{1}{a^2+1}+frac{1}{b^2+2}+frac{1}{c^2+3} = frac{1}{15}(u+v+w)left(frac{1}{u}+frac{1}{v}+frac{1}{w}right). By the AM-HM inequality (Arithmetic Mean - Harmonic Mean inequality), we have frac{1}{15}(u+v+w)left(frac{1}{u}+frac{1}{v}+frac{1}{w}right) geq frac{1}{15} cdot 3sqrt[3]{uvw} cdot frac{3}{sqrt[3]{uvw}} = frac{3}{5}. Equality occurs if and only if u=v=w=5, which corresponds to a^2=4, b^2=3, and c^2=2. Thus, the minimum value of frac{1}{a^2+1}+frac{1}{b^2+2}+frac{1}{c^2+3} is boxed{frac{3}{5}}.

answer:- Calculate the total points scored by the eight other players, who each scored an average of 5 points: [ 8 times 5 = 40 text{ points} ] - Subtract the points scored by the other players from the total team points to find how many points Derek scored: [ 65 - 40 = 25 text{ points} ] Thus, Derek scored boxed{25} points.

answer:1. **Bisectors and exterior angles**: - Exterior angle at B is 180^circ - B, and its bisector yields 90^circ - frac{B}{2}. - Exterior angle at C is 180^circ - C, and its bisector yields 90^circ - frac{C}{2}. - Thus, angle PBC = 90^circ - frac{B}{2} and angle QCB = 90^circ - frac{C}{2}. 2. **Line PQ interaction with PBC and QCB**: - Since PQ forms 30^circ with BC, angles adjacent formed are angle PQB = 30^circ, and angle PQC = 30^circ (as these are consecutive angles across a straight line cutting BC). 3. **Calculating angle PQC**: - In triangle PQC, sum of angles is 180^circ: [ 90^circ - frac{B}{2} + 90^circ - frac{C}{2} + angle PQC = 180^circ ] - Substitute and simplify: [ angle PQC = 180^circ - left(180^circ - frac{B+C}{2}right) = frac{B+C}{2} ] - Realizing B + C = 180^circ - A, substitute for B + C: [ angle PQC = frac{180^circ - A}{2} ] Conclusion: [ angle PQC = frac{180^circ - A{2}} ] The final answer is boxed{textbf{(A)} frac{1}{2}(180-A)}