answer:To solve this problem, we need to evaluate each option step by step: **Option A:** We use the rule of exponents that states a^{m} cdot a^{n} = a^{m+n}. - For a^{2} cdot a^{6}, we have m=2 and n=6. - Therefore, a^{2} cdot a^{6} = a^{2+6} = a^{8}. **Option B:** We use the rule of exponents that states a^{m} div a^{n} = a^{m-n}. - For a^{6} div a^{3}, we have m=6 and n=3. - Therefore, a^{6} div a^{3} = a^{6-3} = a^{3}, not a^{2}. **Option C:** We cannot combine terms with different exponents directly in addition. - 2a^{2} + a^{6} remains as it is because a^{2} and a^{6} are not like terms. - Therefore, it cannot equal 3a^{8}. **Option D:** We use the rule for squaring a product, (ab)^{2} = a^{2}b^{2}. - For (-3a)^{2}, we have (-3)^{2} cdot a^{2} = 9a^{2}, not -9a^{2}. Based on the above evaluations: - Option A is correct because it follows the rule of exponents correctly. - Options B, C, and D are incorrect due to the incorrect application of exponent rules and misunderstanding of term combination and squaring rules. Therefore, the correct option is: boxed{A}.

answer:First, we find the decimal representation of 1/13. Using long division, we find that 1 divided by 13 gives a repeating decimal: 1 div 13 = 0.overline{076923} which repeats every 6 digits. Next, we calculate the 57th digit. We can do this by finding the remainder of 57 divided by the length of the repeating block (6): 57 mod 6 = 3 This means that the 57th digit is the same as the 3rd digit in the repeating sequence. Looking at the sequence 076923, the 3rd digit is boxed{6}.

answer:Extend overline{AB} and overline{CD} to meet at a point F. Then angle AFD = 180^circ - 50^circ - 40^circ = 90^circ. As angle AFD = 90^circ, note that the midpoint of overline{AD}, Q, is the center of the circumcircle of triangle AFD. Similarly, by the properties of trapezoids and their midpoints, midpoint P of overline{BC} is at the center of the circumcircle of triangle BFC. Utilizing the circle properties and the fact that AFD and BFC are right triangles with AFD and BFC being diameters of their respective circumcircles, we find that: [ QF = QD = frac{AD}{2} = 1250, quad PF = PC = frac{BC}{2} = 750. ] Thus the length PQ = QF - PF = 1250 - 750 = boxed{500}. Since overline{BC} parallel overline{AD}, BC and AD are homothetic with respect to point F by a ratio of frac{BC}{AD} = frac{1500}{2500} = frac{3}{5}. Since the homothety carries the midpoint of overline{BC}, P, to the midpoint of overline{AD}, which is Q, then F, P, Q are collinear.

answer:To find out how many drawings were in the first five pages, we need to calculate the number of drawings on each page based on the pattern given: the number of drawings doubles after every page. Let's denote the number of drawings on the first page as D. Page 1: D drawings Page 2: 2D drawings (since the number of drawings doubles) Page 3: 2 * 2D = 4D drawings (doubling again) Page 4: 2 * 4D = 8D drawings (doubling once more) Page 5: 2 * 8D = 16D drawings (doubling for the fifth page) Now, we know that there are 5 drawings on the first page, so D = 5. Let's substitute D with 5 and calculate the total number of drawings in the first five pages: Page 1: 5 drawings Page 2: 2 * 5 = 10 drawings Page 3: 4 * 5 = 20 drawings Page 4: 8 * 5 = 40 drawings Page 5: 16 * 5 = 80 drawings Now, we add the number of drawings from all five pages together: 5 (Page 1) + 10 (Page 2) + 20 (Page 3) + 40 (Page 4) + 80 (Page 5) = 155 drawings So, there were boxed{155} drawings in the first five pages of the art book.