answer:Let the four intersection points be (a,a^2), (b,b^2), (c,c^2), and (d,d^2). Given points are (-3,9), (1,1), and (4,16). Using Vieta's formulas, the sum of the roots a + b + c + d = 0. Given a=-3, b=1, and c=4, we find d = -((-3) + 1 + 4) = -2. Now, calculate the Euclidean distances from the focus (0, frac{1}{4}) to each intersection point: 1. Distance to (-3,9): sqrt{(-3-0)^2 + (9-frac{1}{4})^2} = sqrt{9 + 76.5625} = sqrt{85.5625}. 2. Distance to (1,1): sqrt{(1-0)^2 + (1-frac{1}{4})^2} = sqrt{1 + 0.5625} = sqrt{1.5625}. 3. Distance to (4,16): sqrt{(4-0)^2 + (16-frac{1}{4})^2} = sqrt{16 + 248.0625} = sqrt{264.0625}. 4. Distance to (-2,4): sqrt{(-2-0)^2 + (4-frac{1}{4})^2} = sqrt{4 + 14.0625} = sqrt{18.0625}. Sum of all distances: [ sqrt{85.5625} + sqrt{1.5625} + sqrt{264.0625} + sqrt{18.0625} ] Calculate the exact sums: [ sqrt{85.5625} + sqrt{1.5625} + sqrt{264.0625} + sqrt{18.0625} = 9.25 + 1.25 + 16.25 + 4.25 = boxed{31} ]

answer:The values are (f = 4, g = 5, h = 6, j = 7). Starting similarly: [ (f+g+h+j)^2 = f^2 + g^2 + h^2 + j^2 + 2(fg + fh + fj + gh + gj + hj) ] This implies: [ fg + gh + hj + fj = frac{(f+g+h+j)^2 - f^2 - g^2 - h^2 - j^2}{2} - (fh + gj) ] The task is now to find a configuration that minimizes (fh - gj), which we modify to minimize (fh + gj) for direct application. Testing all pairings: 1. ( fh = 4cdot6 = 24, gj = 5cdot7 = 35 ) then ( fh + gj = 24 + 35 = 59) 2. ( fh = 4cdot 5 = 20, gj = 6cdot7 = 42 ) then ( fh + gj = 20 + 42 = 62) 3. ( fh = 7cdot4 = 28, gj = 5cdot6 = 30 ) then ( fh + gj = 28 + 30 = 58) (minimum) With ( f = 7, g = 5, h = 4, j = 6 ) The minimum (fh + gj = 58). Using all possible sums: [ frac{(4+5+6+7)^2 - 4^2 - 5^2 - 6^2 - 7^2}{2} = frac{484 - 86}{2} = 199 ] [ fg + gh + hj + fj = 199 - 58 = boxed{141} ]

answer:To solve this problem, we need to determine the number N such that the digit 1 appears precisely 15 times and the digit 2 appears precisely 14 times when writing out all integers from 1 to N. 1. **Count the occurrences of digits 1 and 2 from 1 to 9**: - The digits 1 and 2 appear as follows: - 1 appears in 1. - 2 appears in 2. - Each digit 1 and 2 is used 1 time. - Therefore, for the range [1, 9], we have: [ text{Digit 1 used}: 1 text{ time}, quad text{Digit 2 used}: 1 text{ time} ] 2. **Count the occurrences of digits 1 and 2 from 10 to 19**: - The digits 1 and 2 appear as follows: - 10 (1), 11 (1 and 1), 12 (1 and 2), 13 (1), 14 (1), 15 (1), 16 (1), 17 (1), 18 (1), 19 (1). - The digit 1 appears 11 times: [ 1 times 10 = 10 text{ (for tens place)} + 1 text{ (from single place 12)} = 11 text{ times} ] - The digit 2 appears 1 time: [ text{Only in 12} = 1 text{ time} ] 3. **Count the occurrences of digits 1 and 2 from 20 to 29**: - The digits 1 and 2 appear as follows: - 20 (2), 21 (2 and 1), 22 (2 and 2), 23 (2), 24 (2), 25 (2), 26 (2), 27 (2), 28 (2), 29 (2). - The digit 1 appears 1 time: [ text{Only in 21} = 1 text{ time} ] - The digit 2 appears 11 times: [ 2 times 10 = 10 text{ (for tens place) } + 1 text{ (from single place 22)} = 11 text{ times} ] 4. **Count the occurrences of digits 1 and 2 from 30 to 39**: - The digits 1 and 2 appear as follows: - 31 (1), 32 (2), 33, 34, 35, 36, 37, 38, 39. - The digit 1 appears 1 time: [ text{Only in 31} = 1 text{ time} ] - The digit 2 appears 1 time: [ text{Only in 32} = 1 text{ time} ] 5. **Summarize the digit counts up to 39**: - Combined digit counts from ranges [1,9], [10,19], [20,29], and [30,39]: [ text{Digit 1}: 1 (text{[1,9]}) + 11 (text{[10,19]}) + 1 (text{[20,29]}) + 1 (text{[30,39]}) = 14 ] [ text{Digit 2}: 1 (text{[1,9]}) + 1 (text{[10,19]} ) + 11 (text{[20,29]})+ 1 (text{[30,39]})= 14 ] 6. **Count beyond 39 to satisfy the digit 1 condition**: - Check numbers 40 & 41: - 40 does not add any new 1s or 2s. - 41 adds 1 more digit 1. Thus, after writing all integers from 1 to 41: [ text{Digit 1 appears } 14 + 1 = 15 text{ times} ] [ text{Digit 2 appears unchanged } 14 text{ times} ] Since it satisfies the given conditions, the value of N is: [ boxed{41} ]

answer:Let's denote the common difference of the arithmetic sequence as (d), and the (n)th term as (a_n). The first term (a_1) is given as 4. The sum (S_n) of the first (n) terms of an arithmetic sequence can be calculated using the formula: [ S_n = frac{n}{2} (2a_1 + (n - 1)d) ] We are given that the sum of the first 20 terms (S_{20}) is 650, so we can plug in the values we know: [ 650 = frac{20}{2} (2 cdot 4 + (20 - 1)d) ] [ 650 = 10 (8 + 19d) ] [ 650 = 80 + 190d ] [ 570 = 190d ] [ d = frac{570}{190} ] [ d = 3 ] Now that we have the common difference, we can find the second term (a_2) using the formula for the (n)th term of an arithmetic sequence: [ a_n = a_1 + (n - 1)d ] For the second term (a_2): [ a_2 = 4 + (2 - 1) cdot 3 ] [ a_2 = 4 + 3 ] [ a_2 = 7 ] So, the second term of the sequence is boxed{7} .