answer:First, let's calculate the total cost of Claire's daily treats for a week: 1 latte costs 3.75 1 croissant costs 3.50 The total cost for one morning is 3.75 + 3.50 = 7.25 Since she plans to do this every morning for a week (7 days), the total cost for the week is: 7.25/day * 7 days = 50.75 Next, let's calculate the cost of the 5 cookies she plans to buy: 1 cookie costs 1.25 5 cookies cost 1.25 * 5 = 6.25 Now, let's add the cost of the cookies to the weekly cost of her morning treats: 50.75 (weekly treats) + 6.25 (cookies) = 57.00 Finally, we know that after spending on her treats and cookies, she has 43 left on her card. To find the original value of the gift card, we add what she spent to what she has left: 57.00 (spent) + 43.00 (left) = 100.00 Therefore, the value of the gift card was boxed{100.00} .

answer:1. **Case 1: 3 and x+2 are equal.** We have 3 = x + 2. Solving for x, we get: [ x = 3 - 2 = 1 ] Since the third quantity y - 4 must be no greater than the common value 3, we have: [ y - 4 leq 3 implies y leq 7 ] This describes a ray on the line x = 1 starting from (1, 7) and extending downwards indefinitely. 2. **Case 2: 3 and y-4 are equal.** We have 3 = y - 4. Solving for y, we get: [ y = 3 + 4 = 7 ] Since the third quantity x + 2 must be no greater than the common value 3, we have: [ x + 2 leq 3 implies x leq 1 ] This describes a ray on the line y = 7 starting from (1, 7) and extending leftwards indefinitely. 3. **Case 3: x+2 and y-4 are equal.** We have x + 2 = y - 4. Solving for y, we get: [ y = x + 6 ] Since the third quantity 3 must be no less than the common value x + 2, we have: [ 3 leq x + 2 implies x geq 1 ] Also, since 3 leq y - 4, we have: [ 3 leq y - 4 implies y geq 7 ] This describes a ray on the line y = x + 6 starting from (1, 7) and extending upwards indefinitely. 4. **Conclusion:** All three cases describe rays that have a common endpoint at (1, 7). The rays extend from this point in different directions: vertically downwards, horizontally leftwards, and diagonally upwards. Therefore, the set S consists of three rays with a common endpoint. Thus, the correct description for S is boxed{textbf{(E)} text{three rays with a common endpoint}}.

answer:Analysis: This problem tests the properties of logarithms and the application of basic inequalities. Pay attention to the conditions for the equality to hold, which is a basic problem. According to the properties of logarithms, we can get log_2(ab) = 6, so ab = 2^6. By the basic inequality, we can get a + b geq 2sqrt{ab}, and thus find the minimum value of a + b. Solution Enhancement: Given that log_2(a) + log_2(b) = 6, we can rewrite this expression using the product rule of logarithms as log_2(ab) = 6. Then, we can convert the logarithmic equation to exponential form to find that ab = 2^6 = 64. Now, we apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM inequality), which states that the arithmetic mean of two non-negative real numbers is always greater than or equal to the geometric mean of the two numbers, with equality if and only if the two numbers are equal. In this case, the arithmetic mean of a and b is frac{a + b}{2}, while the geometric mean is sqrt{ab}. So, we have frac{a + b}{2} geq sqrt{ab}. Multiplying both sides of the inequality by 2, we get a + b geq 2sqrt{ab}. Since ab = 64, we have a + b geq 2sqrt{64} = 16. The equality holds when a = b. Therefore, the minimum value of a + b is achieved when a = b = sqrt{64} = 8. Hence, the minimum value of a + b is boxed{16}.

answer:Given: In quadrilateral (ABCD), the extension lines of a pair of opposite sides (BA) and (CD) intersect at point (M), and (AD perp BC). We draw diagonals intersecting at (M) and extend lines to meet at points (H, L) on one pair of opposite sides, and compute intersections at point (H', L') on the other pair of opposite sides. We need to prove the following equality: [ frac{1}{MH} + frac{1}{ML} = frac{1}{MH'} + frac{1}{ML'} ] Step-by-Step Solution: 1. **Designation of Intersection Point**: Let the extension lines of (AD) and (BC) intersect at point (O). 2. **Applying Menelaus' Theorem to ( triangle BML ) and ( triangle CML )**: According to Menelaus’ Theorem for (triangle BML) with transversal (AOM): [ frac{BA}{AM} cdot frac{OL}{LB} cdot frac{HM}{MH} = 1 quad Longrightarrow quad frac{BA}{AM} = frac{HL}{MH} cdot frac{OB}{LO} ] Similarly, applying Menelaus’ Theorem for (triangle CML) with transversal (AOM): [ frac{CD}{DM} cdot frac{OL}{LC} cdot frac{HM}{MH} = 1 quad Longrightarrow quad frac{CD}{DM} = frac{HL}{MH} cdot frac{OC}{LO} ] 3. **Combining Results:** Adding the results from the two Menelaus' applications: [ LC cdot frac{BA}{AM} + BL cdot frac{CD}{DM} = frac{HL}{MH} cdot left( frac{OB cdot LC + OC cdot BL}{LO} right) ] 4. **Using Linear Properties**: Notice that: [ OB cdot LC + OC cdot BL = BC cdot LO quad (text{from the collinearity conditions}) ] Thus, [ LC cdot frac{BA}{AM} + BL cdot frac{CD}{DM} = BC cdot frac{HL}{MH} ] 5. **Menelaus’ Application on Other Triangles**: Consider triangles formed by intersections on the other pair of opposite sides. According to similar Menelaus' applications on (triangle LCM) and (triangle LBM): [ BC cdot frac{LL'}{L'M} = BL cdot frac{DC}{MD}, quad BC cdot frac{LH'}{H'M} = LC cdot frac{AB}{AM} ] 6. **Contrasting Intersections:** Substitute these back into the previous equation: [ frac{LL'}{L'M} + frac{LH'}{H'M} = frac{LC cdot AB}{AM cdot BC} + frac{BL cdot CD}{DM cdot BC} ] Simplifying using the proportionality derived from Menelaus as implemented earlier, we get: [ frac{1}{MH} + frac{1}{ML} = frac{1}{MH'} + frac{1}{ML'} ] # Conclusion: [ boxed{frac{1}{MH} + frac{1}{ML} = frac{1}{MH'} + frac{1}{ML'}} ]