answer:1. We start with an equilateral triangle ( ABC ) and identify the points ( M ) and ( N ) as described. Point ( M ) lies on side ( AC ), and point ( N ) is on the extension of ( BC ) such that ( BM = MN ). 2. Draw a line through point ( M ) parallel to side ( BC ) and let this line intersect side ( AB ) at point ( K ). 3. Given that ( BC ) is parallel to ( KM ), the corresponding angles formed are congruent. Specifically: [ angle BKM = angle MCN ] In an equilateral triangle, each internal angle is ( 60^circ ). Since ( BC parallel KM ), the alternate interior angles imply: [ angle BKM = angle MCN = 120^circ ] 4. This means that ( triangle AKM ) is also an equilateral triangle because all angles are ( 60^circ ). 5. Next, denote ( angle KBM = angle CBM = alpha ). Then: [ angle KBM = 60^circ - alpha ] 6. Similarly, since ( BC parallel KM ), we deduce: [ angle KMB = angle NMC ] 7. Given ( BM = MN ) by the problem statement, triangles ( triangle BKM ) and ( triangle MCN ) are congruent by the side-angle-side congruence criterion: [ triangle BKM cong triangle MCN ] 8. From the congruence of these triangles, we can conclude that: [ KM = CN ] 9. Since ( triangle AKM ) is equilateral and ( AK = KM ), it follows that: [ AM = KM ] 10. Thus ( AM = CN ), as required. # Conclusion: [ boxed{AM = CN} ]

answer:Begin with 256 = 4^4 cans. Once recycled: - From 256 cans, make 256 cdot frac{1}{4} = 4^3 = 64 new cans. - Recycle 64 cans to make 64 cdot frac{1}{4} = 4^2 = 16 new cans. - Recycle 16 cans to make 16 cdot frac{1}{4} = 4^1 = 4 new cans. - Finally, recycle these 4 cans to make 4 cdot frac{1}{4} = 4^0 = 1 new can. This continues the process using a geometric series with the first term 64, ratio frac{1}{4}, and four terms. Compute the sum: S = frac{64left(1-left(frac{1}{4}right)^4right)}{1-frac{1}{4}} = frac{64 cdot left(1-frac{1}{256}right)}{0.75} = frac{64 cdot frac{255}{256}}{0.75} = frac{16320}{192} = 85. So the sum is boxed{85}.

answer:To find the average weight of the whole class, we need to find the total weight of all the students in both sections and then divide by the total number of students. First, let's find the total weight of students in section A: Total weight of section A = average weight of section A * number of students in section A Total weight of section A = 40 kg/student * 36 students = 1440 kg Next, let's find the total weight of students in section B: Total weight of section B = average weight of section B * number of students in section B Total weight of section B = 35 kg/student * 44 students = 1540 kg Now, let's find the total weight of all the students in the class: Total weight of the class = Total weight of section A + Total weight of section B Total weight of the class = 1440 kg + 1540 kg = 2980 kg Finally, let's find the total number of students in the class: Total number of students = number of students in section A + number of students in section B Total number of students = 36 students + 44 students = 80 students Now we can find the average weight of the whole class: Average weight of the class = Total weight of the class / Total number of students Average weight of the class = 2980 kg / 80 students = 37.25 kg/student So, the average weight of the whole class is boxed{37.25} kg.

answer:To find the value of the real number a such that the distances from points A(-3,-4) and B(6,3) to the line l: ax+y+1=0 are equal, we proceed as follows: 1. **Express the distances from points to the line using the distance formula**: The distance from a point (x_0, y_0) to a line Ax+By+C=0 is given by dfrac{|Ax_0+By_0+C|}{sqrt{A^2+B^2}}. Applying this formula to our points and line, we get the distances as: - Distance from A to l: dfrac{|-3a-4+1|}{sqrt{a^{2}+1}} - Distance from B to l: dfrac{|6a+3+1|}{sqrt{a^{2}+1}} 2. **Set the distances equal to each other**: [ dfrac{|-3a-4+1|}{sqrt{a^{2}+1}} = dfrac{|6a+3+1|}{sqrt{a^{2}+1}} ] This simplifies to: [ |3a+3| = |6a+4| ] 3. **Solve the equation for a**: We have two cases due to the absolute value: - Case 1: 3a+3 = 6a+4 - Case 2: 3a+3 = -(6a+4) Solving each case: - For Case 1: 3a+3 = 6a+4 simplifies to -3a = 1, giving a = -dfrac{1}{3}. - For Case 2: 3a+3 = -(6a+4) simplifies to 9a = -1, giving a = -dfrac{1}{9}. 4. **Conclusion**: Therefore, the values of a that satisfy the condition are a = -dfrac{1}{3} and a = -dfrac{7}{9}. Thus, the final answer is boxed{a = -dfrac{7}{9} text{ or } -dfrac{1}{3}}.