answer:To show that when a number in decimal notation is divided by 99, the remainder is the same as if the sum of the number formed by partitioning the number into double digits from right to left (the last group might be a single digit) is divided by 99. I. Solution: Consider the general form of a number in decimal notation: [ N = a_{2n+1} cdot 10^{2n+1} + a_{2n} cdot 10^{2n} + a_{2n-1} cdot 10^{2n-1} + a_{2n-2} cdot 10^{2n-2} + ldots + a_3 cdot 10^3 + a_2 cdot 10^2 + a_1 cdot 10 + a_0, ] where the coefficients are single-digit natural numbers, including 0 (i.e., ( a_{2n+1} ) can be 0). The number ( N ) can also be written as: [ begin{aligned} N & = a_{2n+1}(10^{2n+1} - 10 + 10) + a_{2n}(10^{2n} - 1 + 1) + a_{2n-1}(10^{2n-1} - 10 + 10) + ldots & + a_3(10^3 - 10 + 10) + a_2(10^2 - 1 + 1) + a_1 cdot 10 + a_0 & = a_{2n+1}(10^{2n+1} - 10) + a_{2n+1} cdot 10 + a_{2n}(10^{2n} - 1) + a_{2n} + a_{2n-1}(10^{2n-1} - 10) + ldots & + a_3(10^3 - 10) + a_3 cdot 10 + a_2(10^2 - 1) + a_2 + a_1 cdot 10 + a_0 & = (a_{2n+1} cdot 10 + a_{2n})(10^{2n} - 1) + (a_{2n+1} cdot 10 + a_{2n}) + ldots + (a_3 cdot 10 + a_2)(10^2 - 1) + (a_3 cdot 10 + a_2) + (a_1 cdot 10 + a_0). end{aligned} ] Since ((10^{2k} - 1) = 100^k - 1) is divisible by (100 - 1 = 99), dividing (N) by 99 gives the same remainder as the sum of the terms in the square brackets. Thus, the theorem is proven by showing that the sum of these terms, when divided by 99, yields the remainder of (N) divided by 99. blacksquare II. Solution: To verify the theorem, we use the fact that two numbers divided by a third number give the same remainder if and only if their difference is divisible by the third number. Therefore, we will demonstrate that [ left(a_{2n+1} cdot 10^{2n+1} + a_{2n} cdot 10^{2n} + a_{2n-1} cdot 10^{2n-1} + a_{2n-2} cdot 10^{2n-2} + ldots + a_3 cdot 10^3 + a_2 cdot 10^2 + a_1 cdot 10 + a_0right) - left(a_{2n+1} cdot 10 + a_{2n} + a_{2n-1} cdot 10 + a_{2n-2} + ldots + a_3 cdot 10 + a_2 + a_1 cdot 10 + a_0right) ] is divisible by 99. We perform the subtraction: First, pair the terms with their similar indices: [ begin{aligned} & left(a_{2n+1} cdot 10^{2n+1} - a_{2n+1} cdot 10right) + left(a_{2n} cdot 10^{2n} - a_{2n}right) + left(a_{2n-1} cdot 10^{2n-1} - a_{2n-1} cdot 10right) + ldots & + left(a_3 cdot 10^3 - a_3 cdot 10right) + left(a_2 cdot 10^2 - a_2right) end{aligned} ] Factor out the common terms from each pair: [ a_{2n+1} cdot 10 left(10^{2n} - 1right) + a_{2n} left(10^{2n} - 1right) + a_{2n-1} cdot 10 left(10^{2n-2} - 1right) + cdots + a_3 cdot 10 left(10^2 - 1right) + a_2 left(10^2 - 1right). ] It is evident in this form that each term is divisible by ( (10^2 - 1) = 99 ). Hence, the entire expression is divisible by 99. Therefore, the initial number ( N ) divided by 99 has the same remainder as the sum of the terms in square brackets divided by 99. blacksquare

answer:To ensure the function f(x) is well-defined, the following conditions must be met for the variable x: 1. The value under the square root must be non-negative: x - 1 geq 0 2. The argument of the logarithm must be positive: 2 - x > 0 Solving these inequalities: 1. x - 1 geq 0 leads to x geq 1. 2. 2 - x > 0 leads to x < 2. Combining these two conditions, we find that x must lie in the interval where it satisfies both, which is 1 leq x < 2. Therefore, the domain of the function f(x) = sqrt{x-1} + log(2-x) is [1, 2). Finally, we can represent our answer clearly as: [ boxed{[1, 2)} ] This solution reveals the principles necessary for a function's expression to be meaningful, including the consideration of non-negativity for even roots and positivity for logarithmic arguments. The problem tests knowledge of finding the domain of functions, especially logarithmic functions, and highlights the key step of constructing a system of inequalities with respect to x.

answer:1. **Given Information:** We have a triangle with angles alpha, beta, and gamma, such that: [ alpha + beta + gamma = pi ] Additionally, it is given that: [ alpha < beta < gamma ] We need to show that: [ sin 2alpha > sin 2beta > sin 2gamma ] 2. **Using Angle Sum Property of a Triangle:** Since alpha + beta + gamma = pi, it implies that: [ gamma = pi - (alpha + beta) ] Note that gamma is an acute angle because alpha, beta, and gamma are angles of an acute triangle (all angles less than pi/2). 3. **Trigonometric Identity for Sine Difference:** Consider the expression sin 2alpha - sin 2beta. Using the sine difference formula, we get: [ sin 2alpha - sin 2beta = 2 cos left( frac{2alpha + 2beta}{2} right) sin left( frac{2alpha - 2beta}{2} right) ] Simplifying, we have: [ sin 2alpha - sin 2beta = 2 cos (alpha + beta) sin (alpha - beta) ] 4. **Substituting for gamma:** Since gamma = pi - (alpha + beta) and cos(pi - x) = -cos(x) for any angle x, we have: [ cos (alpha + beta) = cos (pi - gamma) = -cos gamma ] Therefore: [ sin 2alpha - sin 2beta = 2 cos (pi - gamma) sin (alpha - beta) = 2 (-cos gamma) sin (alpha - beta) = -2 cos gamma sin (alpha - beta) ] 5. **Considering the Inequalities:** Given alpha < beta < gamma, it follows that alpha - beta < 0, which means sin (alpha - beta) < 0. Since gamma is an acute angle, cos gamma > 0. Hence: [ - 2 cos gamma sin (alpha - beta) > 0 ] This proves that: [ sin 2alpha > sin 2beta ] 6. **Applying Similar Reasoning:** By following the same steps and arguments for sin 2beta and sin 2gamma, we have: [ sin 2beta > sin 2gamma ] 7. **Conclusion:** Combining the results from above, we conclude: [ sin 2alpha > sin 2beta > sin 2gamma ] Therefore: [ boxed{sin 2alpha > sin 2beta > sin 2gamma} ]

answer:The ratio given is 2 parts butter to 5 parts flour to 3 parts sugar. For every 3 parts of sugar, there are 2 parts of butter and 5 parts of flour. Set the number of parts needed for butter and flour based on the sugar requirement: - Sugar requirement is 9 cups, and for each 3 parts of sugar, there should be 2 parts of butter and 5 parts of flour: - For sugar: (3 text{ parts} = 9 text{ cups}), each part equals ( frac{9}{3} = 3 text{ cups}). - Butter needed: (2 times 3 = 6 text{ cups}). - Flour needed: (5 times 3 = 15 text{ cups}). Compute the total cups used: - Total = Butter cups + Flour cups + Sugar cups - Total = (6 + 15 + 9 = boxed{30}).