answer:Solution: (Ⅰ) Since sin ^{2}C+cos ^{2}C=1, we get sin C= frac {2 sqrt {6}}{5}. Thus, sin left(C+ frac {pi}{4}right)=sin Ccdot cos frac {pi}{4}+cos Ccdot sin frac {pi}{4}= frac {2 sqrt {6}}{5}times frac { sqrt {2}}{2}+ frac {1}{5}times frac { sqrt {2}}{2}= frac {4 sqrt {3}+ sqrt {2}}{10}. (Ⅱ) Since overrightarrow{CA}cdot overrightarrow{CB}=|overrightarrow{CA}||overrightarrow{CB}|cos C=1, we have ab=5. Also, a+b= sqrt {37}, so a^{2}+b^{2}=(a+b)^{2}-2ab=27. Therefore, c^{2}=a^{2}+b^{2}-2abcos C=25. Thus, c=5. Therefore, the area of triangle ABC, S_{triangle ABC}= frac {1}{2}absin C= sqrt {6}. So, the answers are boxed{frac {4 sqrt {3}+ sqrt {2}}{10}} for part (Ⅰ), and for part (Ⅱ), the value of side c is boxed{5} and the area of triangle ABC is boxed{sqrt {6}}.

answer:Solution:A. The domain of f(x) = x + 1 is R, while the domain of g(x) = frac{x^2 + x}{x} is {x|x ≠ 0}. As the domains are different, they are not the same function.B. The domain of f(x) = frac{x^2}{(sqrt{x})^2} is (0, +∞), while the domain of g(x) = x is R. As the domains are different, they are not the same function.C. For f(x) = |x|, g(x) = sqrt[n]{x^n} = begin{cases} x & text{n is odd} |x| & text{n is even} end{cases}. As the expressions are different, they are not the same function.D. The domain of f(x) = x is R, and the domain of g(t) = frac{t^3 + t}{t^2 + 1} = t is also R. Both the domains and the expressions are the same, so they are the same function.Therefore, the answer is: boxed{text{D}}.By finding the domain, we can determine that the functions in options A and B are not the same. By looking at the expressions, we can determine that the functions in option C are not the same. Therefore, the only remaining option is D.This question tests the understanding of the definition of a function and the method to determine whether two functions are the same: by checking if both the domain and the expression are the same.

answer:Given that ABCD is a parallelogram, we can use the properties of parallelograms to solve for the angles. Specifically, we know two things:1. Opposite angles in a parallelogram are equal, which means angle B = angle D.2. Adjacent angles in a parallelogram sum up to 180^{circ}, so angle A + angle B = 180^{circ}.Given that angle A is 40^{circ} less than angle B, we can express this as angle B - angle A = 40^{circ}.Now, let's solve for angle B using the information that angle A + angle B = 180^{circ} and angle B - angle A = 40^{circ}.1. From angle A + angle B = 180^{circ} and angle B - angle A = 40^{circ}, we can set up a system of equations: - angle A + angle B = 180^{circ} - angle B - angle A = 40^{circ}2. Adding these two equations, we get: - 2angle B = 220^{circ} - Therefore, angle B = 110^{circ}.3. Substituting angle B = 110^{circ} into angle B - angle A = 40^{circ}, we find: - 110^{circ} - angle A = 40^{circ} - Thus, angle A = 70^{circ}.Since angle C is opposite to angle A in the parallelogram, and opposite angles are equal, we have angle C = angle A = 70^{circ}.Therefore, the degree of angle C is boxed{70^{circ}}.

answer:19. Clearly, if3.6^{circ}1, then letn_{1}=frac{n}{d}, n_{2}=frac{2004}{d}.At this point, we have alpha=frac{n_{1} times 360^{circ}}{n_{2}}.This means that after the pointer rotates n_{2} times, each time rotating by alpha, the pointer will rotate n_{1} full circles and return to its initial position, which contradicts the problem statement.From the above discussion, we know that for any n satisfying 21 leqslant n leqslant 1001 and (n, 2004)=1, there corresponds a possible alpha. Conversely, this is also true. Therefore, the problem becomes finding the number of all n that satisfy the above two conditions.Since 2004=2^{2} times 3 times 167,we have (n, 2004)=1 Leftrightarrow 2 chi_{n}, 3 psi_{n}, 167 uparrow_{n}.Among the positive integers not greater than 1001, the number of positive integers that are not divisible by 2 or 3 isbegin{array}{l}1001-left(left[frac{1001}{2}right]+left[frac{1001}{3}right]-left[frac{1001}{2 times 3}right]right) =1001-(500+333-166)=334 text { (numbers). }end{array}(The symbol [a] represents the greatest integer not exceeding a.)Among them, only 1 times 167 and 5 times 167 are divisible by 167, so the number of n not greater than 1001 and satisfying the conditions is 334-2=332. Removing the 7 numbers 1, 5, 7, 11, 13, 17, 19 that are not greater than 20, we find that the number of n satisfying both conditions is 332-7=325.Therefore, alpha has 325 possible distinct values.

answer:I. The right-hand side segmented decimal fractions in the solution of problem 1302 are converted into the ratios of two natural numbers A / B and C / D, similar to the procedure seen in 1 1. The seventh decimal place of the first segmented decimal fraction marks the beginning of the infinite repetition of the 538461 segment. This is a consequence of the fact that when determining the digits of the A: B ratio up to 6.6538461 (and already subtracting the partial product of 1 with B from the partial dividend), the same remainder R reappeared as it did after determining the digits of the ratio up to 6.6, except that in the latter case, the local value of R is 1 / 10, while in the case of repetition, it is 1 / 10000000 = 1 / 10^{7}. Thus, the two remainder divisions yield the following expressions for the next representation of A:A = B cdot 6.6 + R / 10text{or} quad A = B cdot 6.6538461 + R / 10^{7}Subtracting the first equality from the second, multiplied by 10^{6}, gives the desired form:left(10^{6}-1right) A = B(6653846.1 - 6.6),frac{A}{B} = frac{66538461 - 66}{10 cdot (10^{6} - 1)} = frac{66538395}{9999990} = frac{173}{26}In the simplification, we used the fact that the denominator can be easily factored:999999 = 999 cdot 1001, quad 999 = 27 cdot 37, quad text{and} quad 1001 = 7 cdot 11 cdot 13Thus, the first segmented decimal fraction can be converted to the common fraction 173/26, which can be easily verified.Similarly, the common fraction form of the right-hand side of equation (2) isfrac{C}{D} = frac{5769230 - 5}{9999990} = frac{15}{26}II. Multiply both equations by x y, and add 2 x y to the new first equation, so that the left side of the final form is equal to the square of the left side of the new second equation:begin{gathered}x^{2} + y^{2} = frac{173}{26} x y x + y = frac{15}{26} x y x^{2} + 2 x y + y^{2} = (x + y)^{2} = left(frac{173}{26} + 2right) x y = frac{225}{26} x yend{gathered}Thus, dividing (1b) by (2a) and then the square of (2a) by (1b), and rearranging, we getx + y = 15, quad x y = 26(We used the fact that x neq 0 and y neq 0, otherwise the system of equations would have no meaning.) Therefore, x and y are equal to the roots of the equationz^{2} - 15 z + 26 = 0which are 13 and 2, in any order:x_{1} = 13, quad y_{1} = 2, quad text{and} quad x_{2} = y_{1}, quad y_{2} = x_{1}The value pair indeed satisfies the system of equations.Steiner György (Budapest, Radnóti M. gyak. g. III. o. t.) Baranyai Zsolt (Budapest, Fazekas M. gyak. g. III. o. t.)Remarks. 1. By introducing a new variable u for x / y, after converting the infinite decimal fractions to common fractions, we arrive at the system of equationsu + frac{1}{u} = frac{173}{26}, quad frac{1}{y}left(frac{1}{u} + 1right) = frac{15}{26}, quad x = y uFrom the first, u is 13/2 and 2/13, from the second, y is 2 and 13, and from the third, x is 13 and 2.Varga Gabriella (Szombathely, Savaria g. II. o. t.)2. Most of the solutions chose the method of converting the segmented decimal fractions to common fractions, which, as mentioned in the second remark of problem 1302, proceeds in a way not clarified in the high school curriculum, performing subtraction with numbers written with infinitely many digits as if they were written with finitely many digits. Those who converted the fractions in the manner described above or mentioned this issue received 1 bonus point.[^0][^0]: { }^{1} The editorial board recommended studying the solution of problem 1302 when this exercise was assigned; see K. M. L. 30 (1965) 18. o.

answer:7. C<D<A<B. Since x inleft(frac{pi}{2}, frac{2 pi}{3}right], we know sin (sin x)=cos left[frac{pi}{2}-sin xright]. And frac{sqrt{3}}{2} leqslant sin x<1, we get cos frac{1}{2}(pi- sqrt{3}) leqslant sin (sin x)<cos frac{1}{2}(pi-2).