answer:1. Identify the vertices of the square and their positions after the folds. Let the vertices of the original square be labeled P, Q, R, and S. Alison folds the square along a dashed line which goes through two of these vertices. 2. Let T be the point onto which vertex P is folded. Mark points U and V on the square as shown in the problem diagram. 3. Since the fold along the dashed line creates two congruent triangles P S V and T S V, it follows that: [ P S = T S quad text{and} quad angle P S V = angle T S V ] This is because congruent triangles have equal corresponding sides and angles. 4. Triangle S U T is a right triangle with a right angle at U. From the properties of folding: [ S U = frac{1}{2} S R = frac{1}{2} P S = frac{1}{2} T S ] 5. Given that S U = frac{1}{2} S R, the angle angle T S U can be found using geometry principles (or referred from another exercise): [ angle T S U = 60^circ ] 6. Identify the angles within triangle S U P. Since this triangle has a right angle at U: [ angle U S P = 90^circ ] 7. Since angle P S V = angle T S V and we know angle T S U = 60^circ, the measure of angle T S V is: [ angle T S V = frac{1}{2} (90^circ - 60^circ) = frac{1}{2} cdot 30^circ = 15^circ ] 8. Finally, use the sum of angles in triangle T S V to find alpha. Knowing that the sum of angles in a triangle is 180^circ: [ alpha + 15^circ + 90^circ = 180^circ ] Solve for alpha: [ alpha = 180^circ - 105^circ = 75^circ ] # Conclusion: [ boxed{75} ]

answer:Given m > n, we need to determine which of the following conclusions is correct. **Option A**: frac{m}{5} < frac{n}{5} Given that m > n, dividing both sides by 5, which is a positive number, maintains the inequality direction, so we have: [m > n implies frac{m}{5} > frac{n}{5}] This shows that Option A is incorrect because it suggests the opposite of what is true. **Option B**: m-3 < n-3 Subtracting the same number from both sides of an inequality does not change its direction, so: [m > n implies m-3 > n-3] This indicates that Option B is incorrect as it suggests the opposite of what is true. **Option C**: m+c > n+c Adding the same number to both sides of an inequality does not change its direction, so: [m > n implies m+c > n+c] This confirms that Option C is correct as it maintains the original inequality's direction. **Option D**: -2m > -2n Multiplying both sides of an inequality by a negative number reverses the inequality's direction, so: [m > n implies -2m < -2n] This shows that Option D is incorrect because it suggests the opposite of what is true. Therefore, the correct answer is: boxed{C}.

answer:Given that |2a-3c|=|2b-3c|=frac{2}{3}|d-a|=1 and a neq b, we can analyze the equations step by step. 1. **Equality of Distances**: The equality |2a-3c|=|2b-3c| implies that the distances on the number line between the points representing 2a-3c and 2b-3c are equal. This can be interpreted in two ways: - Either 2a-3c = 2b-3c, which simplifies to 2a = 2b, contradicting a neq b, - Or 2a-3c = -(2b-3c), leading to 2a-3c = -2b+3c, which simplifies to 2a + 2b = 6c or a + b = 3c. 2. **Relation Between a, b, and c**: From a + b = 3c, we can deduce that |2a-3c| = |2a - (a+b)| = |a-b| = 1 and similarly, |2b-3c| = |2b - (a+b)| = |b-a| = 1. This confirms that the distance between a and b on the number line is 1, i.e., a-b = pm 1. 3. **Determining b in Terms of a**: Since a-b = pm 1, we can express b as b = a pm 1. 4. **Distance Between B and D**: The distance between points B and D is given by |d-b|. Given d-a = pm 1.5, we substitute b = a pm 1 into |d-b| to get |d - (a pm 1)|. 5. **Evaluating |d - (a pm 1)|**: - When d-a = 1.5, |d - (a pm 1)| = |1.5 pm 1|, which equals 2.5 or 0.5. - When d-a = -1.5, |d - (a pm 1)| = |-1.5 pm 1|, which also equals 2.5 or 0.5. Therefore, the distance between points B and D can be either 0.5 or 2.5. Thus, the final answer is boxed{0.5 text{ or } 2.5}.

answer:Let x represent Charlie's hourly wage. The additional hours he worked in the second week are 30 - 20 = 10 hours. Therefore, the additional earnings can be defined as 10x, and we know these additional earnings amount to 70. Thus, we have the equation: [ 10x = 70 ] Solving for x gives: [ x = frac{70}{10} = 7 ] To find his total earnings for the first two weeks of July, we calculate: [ 20x + 30x = 50x ] Substituting x = 7: [ 50 times 7 = 350 ] Thus, the total earning for the first two weeks is boxed{350.00}.