answer:We want to find an integer value that can be added to the square of a binomial to obtain y^2 + 10y + 33. We can rewrite the expression as (y^2 + 10y + 25) + 8 = (y + 5)^2 + 8. Therefore, we can add 8 to the square of the binomial (y + 5) to obtain y^2 + 10y + 33. The answer is: 8

answer:One set of 2-dozen paper cups is equal to 2 x 12 = 24 paper cups. So, four sets of 2-dozen paper cups is equal to 4 x 24 = 96 paper cups. If Liezl ended up with 5 damaged cups and 30 unused cups, then the total number of cups used is 96 - 5 - 30 = 61 paper cups. Therefore, 61 paper cups were actually used. 61 The answer is: 61

answer:To solve this problem, we need to determine the value of x, which represents the number of meals Megan plans to have with her remaining eggs. Let's break down the information given: Number of eggs Megan initially had: 12 + 12 = 24 Number of eggs used for omelet: 2 Number of eggs used for cake: 4 Number of eggs given to Megan's aunt: (24 - 2 - 4) / 2 Number of eggs left for Megan's meals: 24 - 2 - 4 - ((24 - 2 - 4) / 2) Number of eggs Megan wants per meal: 3 We can set up the equation as follows: Number of eggs left for Megan's meals / Number of meals = Number of eggs per meal (24 - 2 - 4 - ((24 - 2 - 4) / 2)) / x = 3 Let's simplify and solve for x: (24 - 2 - 4 - ((24 - 2 - 4) / 2)) / 3 = x (18 - ((18) / 2)) / 3 = x (18 - 9) / 3 = x 9 / 3 = x 3 = x The value of x is 3. Megan plans to divide her remaining eggs equally for her next 3 meals. 3 The answer is: 3

answer:Recall that a^{-b} = frac{1}{a^b}. Using this property, we have 27^{-frac{1}{3}} = frac{1}{27^{frac{1}{3}}} = frac{1}{3}, since 27^{frac{1}{3}} = 3. Similarly, 32^{-frac{2}{5}} = frac{1}{32^{frac{2}{5}}} = frac{1}{sqrt[5]{32^2}} = frac{1}{4}, since 32 = 2^5 and sqrt[5]{32^2} = sqrt[5]{2^{10}} = 2^2 = 4. Therefore, 27^{-frac{1}{3}} + 32^{-frac{2}{5}} = frac{1}{3} + frac{1}{4} = frac{4}{12} + frac{3}{12} = boxed{frac{7}{12}}. The answer is: frac{7}{12}