answer:To solve the problem where both the Escher prints and Picasso prints need to be in consecutive blocks, treat each set as a single item. 1. **Grouping Escher and Picasso prints**: Treat the 4 Escher prints as one block and the 3 Picasso prints as another block. This simplifies the problem to arranging these two blocks along with 5 other pieces of art (12 total - 4 Escher - 3 Picasso = 5). 2. **Arranging the blocks and other art**: We now have 7 items to arrange (5 individual pieces + 1 Escher block + 1 Picasso block). The total number of arrangements of these 7 items is 7!. 3. **Arranging within the blocks**: The 4 Escher prints can be arranged among themselves in 4! ways, and the 3 Picasso prints can be arranged in 3! ways. Total favorable arrangements: 7! times 4! times 3! 3. **Calculating total arrangements without restrictions**: Without any restrictions, all 12 pieces can be arranged in 12! ways. 4. **Probability Calculation**: The probability is therefore the ratio of favorable outcomes to possible outcomes. [ P = dfrac{7! times 4! times 3!}{12!} ] Calculating the values: [ 7! = 5040, quad 4! = 24, quad 3! = 6, quad 12! = 479001600 ] [ P = dfrac{5040 times 24 times 6}{479001600} = dfrac{725760}{479001600} = dfrac{1}{660} ] Thus, the probability that all four Escher prints and all three Picasso prints are placed consecutively is boxed{dfrac{1}{660}}.

answer:1. **Determine the value of (Gamma)**: Since (Gamma) is the carry over to the next digit position in the addition, it must be (1). Thus: [ Gamma = 1 ] 2. **Determine the value of (A)**: Given that (A + A) ends in (A), the only digit that satisfies this condition is (0). Therefore: [ A = 0 ] 3. **Considering the absence of carry in the tens place**: Since there is no carry-over in the tens place, we next examine (T + T = M). This implies: [ 2T text{ ends in } M ] Given that (2T) must be even, (M) must also be even. 4. **No carry in the hundreds place**: Similarly, in the hundreds place, if there was a carry, the term (E + E + 1) would end in the (M), but since it doesn't, we consider: [ 2E < 10 ] Allowing the possible values (T) can take to be (2, 3, 4). 5. **Testing possible values of (T)**: - **If (T = 2)**: [ 2 cdot 2 = 4 rightarrow M = 4 ] Hence: [ E + E = 7 + 7 = 14 quad text{(not valid as 7 is reused)} ] - **If (T = 3)**: [ 2 cdot 3 = 6 rightarrow M = 6 ] Hence: [ E = 8, quad E+1 = 9 quad text{with} quad beta = 6 quad text{(conflicts because 6 already used)} ] - **If (T = 4)**: [ 2 cdot 4 = 8 rightarrow M = 8 ] Hence: [ E = 9, quad E+1 = 10 ] This configuration does not result in any contradiction. 6. **Conclusion**: Based on this process, the valid assignment of digits which satisfies the given equation ( ThetaThetaTheta A + betaThetaTheta A = Gammaalphamumualpha ): [ boxed{4940 + 5940 = 10880} ]

answer:To find the percentage increase, you can use the following formula: Percentage Increase = [(New Amount - Original Amount) / Original Amount] * 100 In this case, the original amount is 65 (John's initial weekly earnings), and the new amount is 72 (John's weekly earnings after the raise). So, the calculation would be: Percentage Increase = [(72 - 65) / 65] * 100 Percentage Increase = [7 / 65] * 100 Percentage Increase = 0.10769 * 100 Percentage Increase = 10.769 (rounded to three decimal places) Therefore, the percentage increase in John's weekly earnings is approximately boxed{10.77%} .

answer:Let's denote the number of bags Wendy initially had as ( b ). If Wendy didn't recycle 2 of the bags, she recycled ( b - 2 ) bags. Since she earns 5 points for each bag, the points she earned for recycling ( b - 2 ) bags is ( 5 times (b - 2) ). According to the information given, the points she earned for recycling ( b - 2 ) bags is 45 points. So we can set up the equation: [ 5 times (b - 2) = 45 ] Now, let's solve for ( b ): [ 5b - 10 = 45 ] Add 10 to both sides of the equation: [ 5b = 45 + 10 ] [ 5b = 55 ] Now, divide both sides by 5: [ b = frac{55}{5} ] [ b = 11 ] Wendy initially had boxed{11} bags of cans.