answer:Eugene used all but 16 cards from a deck of 52 cards, so he used 52 - 16 = 36 cards. If Eugene used 75 toothpicks for each card, then he used 36 * 75 = 2700 toothpicks. If a box of toothpicks contains 450 toothpicks, then Eugene used 2700 / 450 = 6 boxes of toothpicks. 6 The answer is: 6

answer:First, we simplify frac{9^2}{3} to get 9^2=81 and 81/3=27. Substituting this into the equation, we now have 12^3=27cdot2^{12p}. We can rewrite 12^3 as (2^2cdot3)^3=2^6cdot3^3. Substituting this into the equation, we now have 2^6cdot3^3=27cdot2^{12p}. We can simplify this further by dividing both sides by 2^6 to get 3^3=27cdot2^{12p-6}. Now, we can simplify the right side to get 3^3=3^3cdot2^{12p-6}. Since the bases are equal, we can equate the exponents to get 12p-6=0. Solving for p, we find p=boxed{frac{1}{2}}. The answer is: frac{1}{2}

answer:First, we evaluate the expression inside the innermost absolute value: |-|{-1 + 1}| - 1 | = |-|{0}| - 1 | = |-0 - 1 | = |-1| = 1. Next, we evaluate the expression inside the outer absolute value: | 1 + 1 | = |2| = 2. Therefore, the value of the given expression is boxed{2}.The answer is: 2

answer:Let s be the side length of square A. Since the perimeter of square A is 24 cm, we have 4s = 24 cm, so s = 6 cm. The area of square A is s^2 = 6^2 = 36 square cm. The area of square B is one-fourth the area of square A, so it is frac{1}{4} cdot 36 = 9 square cm. The side length of square B is sqrt{9} = 3 cm. Therefore, the perimeter of square B is 4 cdot 3 = boxed{12} cm. The answer is: 12