answer:20. 9

answer:By the Cauchy-Schwarz inequality, we haveleft(2 a^{2}+1right)left(frac{1}{2}+lambda^{2}right) geqslant(a+lambda)^{2},where lambda is a positive constant to be determined.Equality holds if and only if 4 a^{2}=frac{1}{lambda^{2}}, i.e., lambda^{2}=frac{1}{4 a^{2}}.Thus, sqrt{1+2 a^{2}} geqslant frac{a+lambda}{sqrt{frac{1}{2}+lambda^{2}}}.Again, by the Cauchy-Schwarz inequality, we haveleft(9 b^{2}+40right)left(frac{1}{9}+frac{1}{40} mu^{2}right) geqslant(b+mu)^{2},where mu is a positive constant to be determined.Equality holds if and only if 81 b^{2}=frac{1600}{mu^{2}}, i.e., mu^{2}=frac{1600}{81 b^{2}}.Thus, sqrt{40+9 b^{2}} geqslant frac{b+mu}{sqrt{frac{1}{9}+frac{1}{40} mu^{2}}}.Therefore, 3 sqrt{1+2 a^{2}}+2 sqrt{40+9 b^{2}}geqslant frac{3(a+lambda)}{sqrt{frac{1}{2}+lambda^{2}}}+frac{2(b+mu)}{sqrt{frac{1}{9}+frac{1}{40} mu^{2}}}.To make the right-hand side of the above inequality a constant, it is necessary and sufficient thatfrac{3}{sqrt{frac{1}{2}+lambda^{2}}}=frac{2}{sqrt{frac{1}{9}+frac{1}{40} mu^{2}}}.Given a>0, b>0, a+b=1, substitutinglambda^{2}=frac{1}{4 a^{2}}, mu^{2}=frac{1600}{81 b^{2}},into equation (1) yieldsbegin{array}{l}frac{3}{sqrt{frac{1}{2}+frac{1}{4 a^{2}}}}=frac{2}{sqrt{frac{1}{9}+frac{40}{81 b^{2}}}} Rightarrow frac{40}{9 b^{2}}=1+frac{1}{(1-b)^{2}} Rightarrow 9 b^{4}-18 b^{3}-22 b^{2}+80 b-40=0 Rightarrow(b+2)(3 b-2)left(3 b^{2}-10 b+10right)=0 Rightarrow b=frac{2}{3}, a=frac{1}{3}, lambda=frac{3}{2}, mu=frac{20}{3} .end{array}Therefore, when a=frac{1}{3}, b=frac{2}{3},E(a, b)=3 sqrt{1+2 a^{2}}+2 sqrt{40+9 b^{2}}achieves its minimum value 5 sqrt{11}.

answer:If n is an odd number, then the number 2^n +105 gives a remainder of 2 when divided by 3. Since a number that is the square of an integer cannot give a remainder of 2 when divided by 3, the number n satisfying the conditions of the problem must be even. Let's assume that n =2k for some non-negative integer k . We want to solve the equation 2^{2k} +105 = m^2 in non-negative integers k and m . We rewrite this equation as (m-2^k)(m+2^k) = 105 . From the obtained equation, it follows that m-2^k >0 . Of course, the inequality m+2^k > m-2^k is also satisfied. Since 105=3cdot 5cdot 7 , finding non-negative integers m and k that satisfy the relation (m-2^k)(m+2^k)=105 reduces to solving four systems of equations:By subtracting the first equation from the second, we conclude that the number 2 cdot 2^k =2^{k+1} equals 8, 16, 32, or 104. From this, we obtain three possible values for k ; they are: 2, 3, or 4. Thus, the possible values of the number n are 4, 6, or 8. We directly check that for the three obtained values of n , the number 2^n +105 is the square of an integer:直接检查结果如下：- 当 n = 4 时, 2^4 + 105 = 16 + 105 = 121 = 11^2 .- 当 n = 6 时, 2^6 + 105 = 64 + 105 = 169 = 13^2 .- 当 n = 8 时, 2^8 + 105 = 256 + 105 = 361 = 19^2 .Directly checking the results:- For n = 4 , 2^4 + 105 = 16 + 105 = 121 = 11^2 .- For n = 6 , 2^6 + 105 = 64 + 105 = 169 = 13^2 .- For n = 8 , 2^8 + 105 = 256 + 105 = 361 = 19^2 .

answer:1. Volume of One Pentagonal Prism: - The end of each prism is a combination of a rectangle and an isosceles triangle. - The rectangle has dimensions (12 text{ ft} times 7 text{ ft}). - The isosceles triangle has sides (10 text{ ft}, 10 text{ ft}, 12 text{ ft}).2. Area of the Isosceles Triangle: - Using Heron's formula to find the area of the triangle: [ s = frac{10 + 10 + 12}{2} = 16 text{ ft} ] [ text{Area} = sqrt{s(s-10)(s-10)(s-12)} = sqrt{16 times 6 times 6 times 4} = sqrt{2304} = 48 text{ ft}^2 ]3. Volume of the Top Part of the Prism: - The top part is a triangular prism with base area (48 text{ ft}^2) and length (30 text{ ft}): [ text{Volume}_{text{top}} = 48 times 30 = 1440 text{ ft}^3 ]4. Volume of the Bottom Part of the Prism: - The bottom part is a rectangular prism with dimensions (12 text{ ft} times 7 text{ ft} times 30 text{ ft}): [ text{Volume}_{text{bottom}} = 12 times 7 times 30 = 2520 text{ ft}^3 ]5. Total Volume of One Pentagonal Prism: [ text{Volume}_{text{prism}} = 1440 + 2520 = 3960 text{ ft}^3 ]6. Volume of Two Pentagonal Prisms: [ text{Volume}_{text{two prisms}} = 2 times 3960 = 7920 text{ ft}^3 ]7. Volume of the Common Center: - The common center is a combination of a pyramid and a rectangular box. - The base of the pyramid is a square with side (12 text{ ft}) and height (8 text{ ft}): [ text{Volume}_{text{pyramid}} = frac{1}{3} times 12 times 12 times 8 = frac{1152}{3} = 384 text{ ft}^3 ] - The rectangular box has dimensions (12 text{ ft} times 12 text{ ft} times 7 text{ ft}): [ text{Volume}_{text{box}} = 12 times 12 times 7 = 1008 text{ ft}^3 ]8. Total Volume of the Common Center: [ text{Volume}_{text{common center}} = 384 + 1008 = 1392 text{ ft}^3 ]9. Final Volume of the Barn: [ text{Volume}_{text{barn}} = 7920 - 1392 = 6528 text{ ft}^3 ]The final answer is (boxed{6528 text{ ft}^3}).

answer:[Solution] Transform the given equation intoa cos ^{2} x+c=-b cos x .Square both sides to getbegin{array}{l}a^{2} cos ^{4} x+2 a cos ^{2} x cdot c+c^{2}=b^{2} cos ^{2} x, a^{2} cos ^{4} x+left(2 a c-b^{2}right) cos ^{2} x+c^{2}=0 .end{array}Substitute cos ^{2} x=frac{1+cos 2 x}{2} into equation (1) to geta^{2}left(frac{1+cos 2 x}{2}right)^{2}+left(2 a c-b^{2}right) cdot frac{1+cos 2 x}{2}+c^{2}=0 .Simplify to geta^{2} cos ^{2} 2 x+2left(a^{2}+2 a c-b^{2}right) cos 2 x+(a+2 c)^{2}-2 b^{2}=0 .This is a quadratic equation in terms of cos 2 x.For the case where a=4, b=2, c=-1,The given equation is4 cos ^{2} x+2 cos x-1=0The new equation is4 cos ^{2} 2 x+2 cos 2 x-1=0 .Thus, these two equations are quadratic equations with identical coefficients, where the first one is in terms of cos x and the second one is in terms of cos 2 x.

answer: Part (a)1. Arrange Marbles on the Diagonal: - Place each of the 2024 marbles on one of the two main diagonals of the (2024 times 2024) grid. For example, place a marble at each cell ((i, i)) for (i = 1, 2, ldots, 2024).2. Verify the Condition: - Each marble is placed such that no two marbles are in neighboring cells. Neighboring cells share an edge, but in this arrangement, each marble is only diagonally adjacent to another marble, not sharing an edge.3. Check the Movement Condition: - Moving any marble to a neighboring cell (either up, down, left, or right) will place it in a cell that is already occupied by another marble, violating the condition that no two marbles are in neighboring cells.Thus, the arrangement satisfies the conditions, and moving any marble will violate the conditions. Part (b)1. Claim: (k = 2023) is the Largest Value: - We need to show that for (k = 2023), there exists an arrangement where moving any marble to a neighboring cell will still satisfy the conditions.2. Assume (k = 2024 - n) Marbles: - Let (n geq 1). We need to show that there is always a way to move a marble if (k < 2024).3. Free Rows and Columns: - Since there are fewer than 2024 marbles, there will be rows and columns without any marbles, called free rows and columns. - Let (p) be the number of free columns and (q) be the number of free rows. Without loss of generality, assume (p geq q).4. Claim 1: No Two Free Rows are Adjacent: - If there were two adjacent free rows, there must be a marble adjacent to this block. Moving this marble into the block would create a valid arrangement, contradicting the assumption.5. Claim 2: The Leftmost Column is Not Free: - If the leftmost column were free, moving a marble from the next column to the leftmost column would create a valid arrangement, contradicting the assumption.6. Marbles in Non-Free Columns: - For each free column, the column immediately to the left contains a marble. If a marble is at ((x, y)), there must be another marble at ((x+2, y)) to prevent moving the marble to ((x+1, y)).7. Counting Marbles and Rows: - There are (p) free columns, so there are at least (p) pairs of marbles with a distance vector ((2, 0)). These pairs use up (p) rows. - The remaining (2024 - n - 2p) marbles can use at most (2024 - n - 2p) rows. - Therefore, at most (p + (2024 - n - 2p) = 2024 - n - p) rows have marbles, implying at least (p + n) rows are free.8. Contradiction: - Since (p geq q), the number of free rows (q) must be at least (p + n), which is a contradiction.Thus, the largest value of (k) such that for all arrangements of (k) marbles, we can move one marble to a neighboring cell and still satisfy the conditions is (k = 2023).The final answer is ( boxed{ k = 2023 } )