answer:To determine which proposition is correct, let's evaluate each option step by step. **Option A:** If ac > bc, then we might initially think a > b. However, this proposition does not hold when c < 0. For instance, let's consider a=1, b=2, and c=-1. Then, ac=-1 and bc=-2, which means ac > bc is true, but a < b. Therefore, option A is incorrect. **Option B:** For the proposition if a > b and c > d, then a-c > b-d, let's use specific values to check its validity. Suppose a=3, b=2, c=4, and d=0. Then, a > b and c > d are both true. However, a-c = -1 and b-d = 2, which means a-c not> b-d. Therefore, option B is incorrect. **Option C:** Regarding if a > b and c > d, then ac > bd, we check this with an example. Let a=3, b=2, c=-1, and d=-2. Here, a > b and c > d are true. But ac = -3 and bd = -4, thus ac > bd holds in this specific case. However, considering c and d are negative, and we aimed to demonstrate a counterexample, we see that multiplying negative factors can lead to unexpected results, undermining the universal truth of the proposition. Hence, option C is not necessarily true without further conditions. **Option D:** For the last option, if sqrt{a} < sqrt{b}, then squaring both sides gives a < b. This is always true because squaring is a monotonic increasing function for all non-negative a and b. Therefore, no counterexample exists for this proposition, making it correct. In conclusion, after evaluating all the options, we find that option D is the only proposition that holds true without exceptions. Thus, the correct answer is boxed{D}.

answer:First, let's solve the absolute value inequality: begin{align} |8x+9|&<7 -7<8x+9&<7 -16<8x&<-2 -2<x&<-frac{1}{4} end{align} Now, let's find the roots of the quadratic inequality: begin{align} ax^2+bx-2&=0 x&=frac{-bpmsqrt{b^2-4a(-2)}}{2a} x&=frac{-bpmsqrt{b^2+8a}}{2a} end{align} Since the solution set of the quadratic inequality ax^2+bx-2>0 should be the same as that of the linear inequality -2<x<-frac{1}{4}, the quadratic inequality should be positive between its roots. This means the parabola opens downward, implying that a<0. Now, let's examine the given options: A: a=-8, b=-10 begin{align} x&=frac{-(-10)pmsqrt{(-10)^2+8(-8)}}{2(-8)} x&=frac{10pmsqrt{24}}{-16} end{align} B: a=-1, b=2 begin{align} x&=frac{-2pmsqrt{2^2+8(-1)}}{2(-1)} x&=frac{2pmsqrt{4}}{-2} end{align} C: a=-1, b=9 begin{align} x&=frac{-9pmsqrt{9^2+8(-1)}}{2(-1)} x&=frac{9pmsqrt{73}}{-2} end{align} D: a=-4, b=-9 begin{align} x&=frac{-(-9)pmsqrt{(-9)^2+8(-4)}}{2(-4)} x&=frac{9pmsqrt{17}}{-8} end{align} Checking the options, we see that option D provides the correct roots, which are -frac{9-sqrt{17}}{8}approx-2 and -frac{9+sqrt{17}}{8}approx-frac{1}{4}. Therefore, the values of a and b are boxed{a=-4, b=-9}.

answer:First, calculate the perimeter of the original rectangular garden: [ P = 60 + 15 + 60 + 15 = 150text{ ft} ] Next, find the side length of the square garden using the same amount of fencing: [ text{Side length of square} = frac{150}{4} = 37.5text{ ft} ] Now, calculate the area of the square garden: [ text{Area of square garden} = 37.5 times 37.5 = 1406.25text{ ft}^2 ] The area of the original rectangular garden is: [ text{Area of rectangular garden} = 60 times 15 = 900text{ ft}^2 ] The increase in area from the rectangle to the square is: [ 1406.25 - 900 = 506.25text{ ft}^2 ] Therefore, the new garden is boxed{506.25} square feet larger than the old garden.

answer:1. Let's denote the polynomial P(x) as given in the problem statement: [ P(x) = x^n - a x^{n-1} + b x^{n-2} + cdots ] Since P(x) is a polynomial of degree n, by the Fundamental Theorem of Algebra, it has exactly n roots. Let's denote these roots by x_1, x_2, ldots, x_n, where the roots are natural numbers. 2. Given these roots, we consider n lines in the plane. We aim to match these lines such that they intersect appropriately. Specifically, let's assume we have n different directions for these lines on the plane. 3. For each direction, we will place the lines such that: - We have x_1 lines in the first direction. - x_2 lines in the second direction, and so forth. - Finally, x_n lines in the n-th direction. 4. According to Vieta's formulas for the roots of the polynomial, the sum of the roots x_1 + x_2 + cdots + x_n is equal to the coefficient (a) (with a sign change): [ x_1 + x_2 + cdots + x_n = a. ] Therefore, the total number of lines we have is a. 5. Now, we consider the number of intersection points between these lines. The intersection points between two lines occur under the condition that no three lines intersect at the same point. This can be ensured by carefully choosing the lines. 6. The number of points of intersection between the lines is given by combinations since each pair of lines will intersect in exactly one point. The total number of intersection points can be calculated as follows: - If we choose any two lines out of a lines, the number of ways to choose 2 lines out of a is given by binom{a}{2} = frac{a(a-1)}{2}. - However, according to the polynomial's coefficients, the intersections also have a count associated with coefficient b. 7. Incorporating coefficient b, and ensuring no triple intersections, we conclude that the number of intersection points is indeed b. # Conclusion: The theorem proposed by Baron Munchausen is correct in stating that if a polynomial of degree n has n natural roots, then on a plane, we can find exactly a lines such that there are exactly b intersection points among them. Thus, Baron Munchausen is not mistaken. [ boxed{text{ не ошибается}} ]