answer:1. **Assume the graph ( G ) is connected.** - Given the number of edges in the graph is ( frac{1}{2}(n^2 - 3n + 4) ). 2. **Check the condition for the graph to have a cycle:** - For a connected graph with ( n ) vertices, the minimum number of edges required to ensure connectivity is ( n-1 ). - Therefore, we need: [ frac{1}{2}(n^2 - 3n + 4) > n - 1 ] - Simplify the inequality: [ n^2 - 3n + 4 > 2n - 2 ] [ n^2 - 5n + 6 > 0 ] [ (n-2)(n-3) > 0 ] - This inequality holds for ( n > 3 ). Therefore, if ( G ) has a cycle, then ( n geq 4 ). 3. **Consider the removal of an edge that disconnects the graph:** - Suppose removing an edge from ( G ) results in a disconnected graph ( G' ) with connected components ( G_1 ) and ( G_2 ). - Let ( |G_1| = n_1 ) and ( |G_2| = n_2 ) with ( n_1 + n_2 = n ). 4. **Calculate the number of edges in ( G' ):** - The number of edges in ( G' ) is: [ frac{1}{2}(n^2 - 3n + 4) - 1 = frac{1}{2}(n^2 - 3n + 2) ] - The number of edges in ( G' ) must be less than or equal to the sum of the edges in ( G_1 ) and ( G_2 ): [ frac{1}{2}(n_1(n_1 - 1) + n_2(n_2 - 1)) ] 5. **Set up the inequality:** - We have: [ frac{1}{2}(n^2 - 3n + 2) leq frac{1}{2}(n_1(n_1 - 1) + n_2(n_2 - 1)) ] - Simplify: [ n^2 - 3n + 2 leq n_1^2 - n_1 + n_2^2 - n_2 ] - Substitute ( n_2 = n - n_1 ): [ n^2 - 3n + 2 leq n_1^2 - n_1 + (n - n_1)^2 - (n - n_1) ] [ n^2 - 3n + 2 leq n_1^2 - n_1 + n^2 - 2nn_1 + n_1^2 - n + n_1 ] [ n^2 - 3n + 2 leq 2n_1^2 - 2nn_1 + n^2 - n ] [ -3n + 2 leq 2n_1^2 - 2nn_1 - n ] [ 2n_1^2 - 2nn_1 + 2n - 2 geq 0 ] [ 2(n_1^2 - nn_1 + n - 1) geq 0 ] 6. **Solve the inequality:** - The inequality ( n_1(n_2 - 1) leq 0 ) implies that either ( n_1 = 1 ) or ( n_2 = 1 ). - Without loss of generality, assume ( n_2 = 1 ). This means there is a vertex in ( G ) with degree exactly 1. 7. **Determine the maximum length ( k ) of a circuit:** - If there is a vertex with degree 1, the longest possible cycle in ( G ) can have at most ( n-1 ) vertices. - Therefore, the greatest possible length ( k ) of a circuit in such a graph is ( n-1 ). The final answer is ( boxed{n-1} ).

answer:Let's denote the principal amount as P and the annual interest rate as R (in percentage). The formula for simple interest (SI) is: SI = P * R * T / 100 where T is the time in years. Given that the simple interest for 2 years is Rs. 56, we can write: 56 = P * R * 2 / 100 28 = P * R / 100 (1) The formula for compound interest (CI) for 2 years is: CI = P * (1 + R/100)^2 - P Given that the compound interest for 2 years is Rs. 57.40, we can write: 57.40 = P * (1 + R/100)^2 - P 57.40 = P * [(1 + R/100)^2 - 1] (2) Now, let's express (1 + R/100)^2 as a binomial expansion: (1 + R/100)^2 = 1 + 2*(R/100) + (R/100)^2 Substituting this back into equation (2), we get: 57.40 = P * [1 + 2*(R/100) + (R/100)^2 - 1] 57.40 = P * [2*(R/100) + (R/100)^2] We know from equation (1) that P * R / 100 = 28, so we can substitute this into the equation above: 57.40 = 2 * 28 + P * (R/100)^2 57.40 = 56 + P * (R^2/10000) Now, let's subtract the simple interest (56) from the compound interest (57.40) to find the difference, which is due to the compound interest on the first year's interest: 57.40 - 56 = P * (R^2/10000) 1.40 = P * (R^2/10000) We already know that P * R / 100 = 28, so we can find P by rearranging this equation: P = 28 * 100 / R Now, let's substitute this value of P into the equation for the difference due to compound interest: 1.40 = (28 * 100 / R) * (R^2/10000) 1.40 = (28 * R^2) / (R * 100) 1.40 = 28 * R / 100 1.40 * 100 = 28 * R 140 = 28 * R R = 140 / 28 R = 5 Therefore, the annual interest rate is boxed{5%} .

answer:The form of scientific notation is a times 10^n, where 1 leq |a| < 10, and n is an integer. To determine the value of n, we need to see how many places the decimal point has moved to turn the original number into a. The absolute value of n is the same as the number of places the decimal point has moved. When the absolute value of the original number is greater than 1, n is positive; when the absolute value of the original number is less than 1, n is negative. Therefore, 16,300,000,000 = 1.63 times 10^{10}. So, the number of viewers can be represented in scientific notation as boxed{1.63 times 10^{10}} people.

answer:To find the number of integer solutions to the equation (2^{2x} - 3^{2y} = 55), we start by factoring the left-hand side: 1. Rewrite the equation: [ 2^{2x} - 3^{2y} = 55 ] 2. Notice that (2^{2x} = (2^x)^2) and (3^{2y} = (3^y)^2). Let (a = 2^x) and (b = 3^y). The equation becomes: [ a^2 - b^2 = 55 ] 3. Factor the difference of squares: [ (a - b)(a + b) = 55 ] 4. Consider the factor pairs of 55: [ 55 = 1 times 55, quad 55 = 5 times 11 ] 5. Analyze each factor pair to find integer solutions for (a) and (b): - **First pair: (a - b = 1) and (a + b = 55)** [ begin{cases} a - b = 1 a + b = 55 end{cases} ] Add the equations: [ 2a = 56 implies a = 28 ] Substitute (a = 28) into (a - b = 1): [ 28 - b = 1 implies b = 27 ] Check if (a = 28) and (b = 27) correspond to powers of 2 and 3: [ 2^x = 28 quad text{(not a power of 2)} ] Therefore, this pair does not yield integer solutions for (x) and (y). - **Second pair: (a - b = 5) and (a + b = 11)** [ begin{cases} a - b = 5 a + b = 11 end{cases} ] Add the equations: [ 2a = 16 implies a = 8 ] Substitute (a = 8) into (a - b = 5): [ 8 - b = 5 implies b = 3 ] Check if (a = 8) and (b = 3) correspond to powers of 2 and 3: [ 2^x = 8 implies x = 3 3^y = 3 implies y = 1 ] Therefore, (x = 3) and (y = 1) is a valid solution. 6. Since the first pair did not yield integer solutions and the second pair did, we conclude that there is exactly one solution. The final answer is (boxed{1})