BERT

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question:Khali now has an additional portion of the sidewalk to shovel. The main sidewalk is 30 feet long and 3 feet wide, while the new portion is a square of side 4 feet and has snow 1 foot deep. The main portion still has the snow 0.5 feet deep. How many cubic feet of snow does Khali have to shovel off in total from both portions?

answer:1. **Calculating the volume of snow on the main sidewalk**: The volume of the snow on the main portion can be calculated using the volume formula for a rectangular prism: [ text{Volume}_{text{main}} = text{Length} times text{Width} times text{Height} = 30 text{ ft} times 3 text{ ft} times 0.5 text{ ft} = 45 text{ cubic feet} ] 2. **Calculating the volume of snow on the additional square portion**: The volume of the snow on the square portion is calculated similarly: [ text{Volume}_{text{square}} = text{Side} times text{Side} times text{Height} = 4 text{ ft} times 4 text{ ft} times 1 text{ ft} = 16 text{ cubic feet} ] 3. **Total volume of snow to shovel**: The total volume of snow Khali has to shovel is the sum of the volumes from the two portions: [ text{Total Volume} = text{Volume}_{text{main}} + text{Volume}_{text{square}} = 45 text{ cubic feet} + 16 text{ cubic feet} = boxed{61 text{ cubic feet}} ] Conclusion: The total amount of snow Khali needs to shovel is 61 cubic feet.

question:What is the greatest common divisor of 462 and 330?

answer:To solve the problem, we will first identify the factors of 462 and 330. 1. **Factors of 462**: - Prime factorization of 462: ( 462 = 2 times 3 times 7 times 11 ) - Factors: 1, 2, 3, 6, 7, 14, 21, 42, 11, 22, 33, 66, 77, 154, 231, 462 2. **Factors of 330**: - Prime factorization of 330: ( 330 = 2 times 3 times 5 times 11 ) - Factors: 1, 2, 3, 6, 5, 10, 15, 30, 11, 22, 33, 66, 55, 110, 165, 330 3. **Common Factors**: - Comparing the lists: 1, 2, 3, 6, 11, 22, 33, 66 4. **Largest Common Factor**: - The greatest number in the list of common factors is 66. Therefore, the greatest common divisor of 462 and 330 is boxed{66}.

question:The negation of the proposition There exists x in (0, +infty), such that ln(x) = x 1 is ( ) A: For any x in (0, +infty), ln(x) neq x 1 B: For any x notin (0, +infty), ln(x) = x 1 C: There exists x in (0, +infty), ln(x) neq x 1 D: There exists x notin (0, +infty), ln(x) = x 1

answer:Since the negation of an existential proposition is a universal proposition, the negation of "There exists x in (0, +infty), such that ln(x) = x - 1" is: For any x in (0, +infty), ln(x) neq x - 1, Therefore, the correct choice is: boxed{A}. By utilizing the negation of an existential proposition as a universal proposition, the result can be directly written. This question examines the relationship between the negation of existential and universal propositions, testing the application of basic knowledge.

question:The function f(x) is an odd function defined on the real numbers mathbb{R}, and when x in (0, +infty), f(x) = 2^x. Calculate f(1).

answer:Since f(x) is an odd function defined on mathbb{R}, this means for any x in its domain, f(-x) = -f(x). Given that f(x) = 2^x for x in (0, +infty), we can determine the value of f(1) directly: f(1) = 2^1 = 2. Using the property of odd functions, for x=1, we find f(-1) = -f(1) = -2. Therefore, the value of f(-1) is boxed{-2}.