I'm unable to access specific content from MathBits or any other external sources in real-time. However, for basic caching, the answer typically involves understanding how data is temporarily stored for quick access to improve efficiency and reduce load times. If you have specific options or concepts from box 2 that you want to discuss, feel free to share, and I can provide insights or explanations based on general caching principles.
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1) -18 2) -22 3) -96 4) -8 5) -85 6) -16 Final answer= 8296
Nine, assuming the two boxes are considered different: 8 in the first box 7 in the first box, 1 in the second 6 in the first box, 2 in the second 5 in the first box, 3 in the second 4 in each box 3 in the first box, 5 in the second 2 in the first box, 6 in the second 1 in the first box, 7 in the second 8 in the second box. If the boxes are interchangable, there are only five: 8 in one box 7 in one box, 1 in the other 6 in one box, 2 in the other 5 in one box, 3 in the other 4 in each box
6 is your answer because the box is 3 lengths and 2 widths and its basicly a 3 by 2 box.
384
For a box with equal dimensions on all sides, the surface area, S, is calculated as follows: S = 6*a^2 where a is the length of the side of the box For a box with different dimensions on the sides, the surface area, S, is calculated as follows: S = 2*a*b + 2*a*c + 2*b*c where a, b, c are the lengths of the different sides of the box