answer:First, note the coordinates of the triangle vertices: - Vertex A at (2, 2) - Vertex B at (8, 2) - Vertex C at (4, 10) From the given coordinates, AB is the base of the triangle which lies parallel to the x-axis. The length of AB can be calculated as: [ |B_x - A_x| = |8 - 2| = 6 ] The height of the triangle is the perpendicular distance from C to line AB, which is the difference in the y-coordinates of C and A (or B, as they lie on the same horizontal line): [ |C_y - A_y| = |10 - 2| = 8 ] Using the formula for the area of a triangle frac{1}{2} times text{base} times text{height}, we get: [ text{Area} = frac{1}{2} times 6 times 8 = 24 ] Thus, the area of the triangle is (boxed{24}) square units.

answer:Given (a, b in mathbb{N}) such that: [ ab + 2 = a^3 + 2b ] 1. Rearrange the given equation: [ ab + 2 = a^3 + 2b implies a^3 - ab = 2b - 2 ] 2. Factorize and reorganize: [ a^3 - ab - 2b + 2 = 0 implies a^3 - ab = 2b - 2 implies a^3 - 2 = b(a - 2) ] 3. We notice that: [ b(a - 2) = a^3 - 2 ] implies (a neq 2) because (2^3 - 2 = 6 neq 0). 4. Since (a neq 2), we have (a - 2 mid a^3 - 2). 5. Use Theorem: If (d mid x) and (d mid y), then (d mid (x + y)). So, let's apply factoring: [ a^3 - 2 = (a^3 - 8) + 6 implies a^3 - 8 = (a - 2)(a^2 + 2a + 4) ] 6. Therefore: [ a - 2 mid 6 ] This implies (a - 2) must be a divisor of 6. The divisors of 6 are ({-6, -3, -2, -1, 1, 2, 3, 6}). 7. Consequently: [ a - 2 in {-6, -3, -2, -1, 1, 2, 3, 6} ] which gives: [ a in {-4, -1, 0, 1, 3, 4, 5, 8} ] Since (a in mathbb{N}) (non-negative integers), we have: [ a in {0, 1, 3, 4, 5, 8} ] 8. Now, find (b) for each (a): - For (a = 0): [ 0^3 - 2 = b(0 - 2) implies -2 = -2b implies b = 1 ] - For (a = 1): [ 1^3 - 2 = b(1 - 2) implies -1 = -b implies b = 1 ] - For (a = 3): [ 3^3 - 2 = b(3 - 2) implies 27 - 2 = b(1) implies 25 = b implies b = 25 ] - For (a = 4): [ 4^3 - 2 = b(4 - 2) implies 64 - 2 = b(2) implies 62 = 2b implies b = 31 ] - For (a = 5): [ 5^3 - 2 = b(5 - 2) implies 125 - 2 = b(3) implies 123 = 3b implies b = 41 ] - For (a = 8): [ 8^3 - 2 = b(8 - 2) implies 512 - 2 = b(6) implies 510 = 6b implies b = 85 ] 9. Verify the solutions: - For ((a, b) = (0, 1)): [ 0 cdot 1 + 2 = 0^3 + 2 cdot 1 implies 2 = 2 quad text{correct!} ] - For ((a, b) = (1, 1)): [ 1 cdot 1 + 2 = 1^3 + 2 cdot 1 implies 3 = 3 quad text{correct!} ] - For ((a, b) = (3, 25)): [ 3 cdot 25 + 2 = 3^3 + 2 cdot 25 implies 75 + 2 = 27 + 50 implies 77 = 77 quad text{correct!} ] - For ((a, b) = (4, 31)): [ 4 cdot 31 + 2 = 4^3 + 2 cdot 31 implies 124 + 2 = 64 + 62 implies 126 = 126 quad text{correct!} ] - For ((a, b) = (5, 41)): [ 5 cdot 41 + 2 = 5^3 + 2 cdot 41 implies 205 + 2 = 125 + 82 implies 207 = 207 quad text{correct!} ] - For ((a, b) = (8, 85)): [ 8 cdot 85 + 2 = 8^3 + 2 cdot 85 implies 680 + 2 = 512 + 170 implies 682 = 682 quad text{correct!} ] These verifications confirm the pairs are correct. Thus, the pairs of natural numbers ((a, b)) are: [ boxed{{(0,1),(1,1),(3,25),(4,31),(5,41),(8,85)}} ]

answer:Step 1: When a < 0, A=varnothing. It is obvious that Acap B=varnothing holds. Step 2: When ageqslant 0, A={x,|,2-aleqslant xleqslant 2+a}, and B={x,|,xleqslant 1 text{ or } xgeqslant 4}. From Acap B=varnothing, we get begin{cases}2-a > 1 2+a < 4 ageqslant 0end{cases}, which solves to 0leqslant a < 1. In summary, the range of a is boxed{(-infty,1)}.

answer:Since {a_n} is an arithmetic sequence, and S_{15} > 0, S_{16} < 0, We can deduce that a_8 > 0, a_8 + a_9 < 0 Thus, a_9 < 0, So, among {frac{S_1}{a_1}, frac{S_2}{a_2}, ...}, the first 8 terms are positive, and terms 9 to 15 are negative. Furthermore, among the first 8 terms, the numerators are increasing while the denominators are decreasing. Therefore, the maximum value among {frac{S_1}{a_1}, frac{S_2}{a_2}, ..., frac{S_{15}}{a_{15}}} is boxed{frac{S_8}{a_8}}.