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question:Given i is the imaginary unit, and complex numbers z_1=3bi, z_2=12i, if frac{z_1}{z_2} is a real number, then the value of the real number b is ( ) A: 6 B: 6 C: 0 D: frac{1}{6}
answer:Solution: frac{z_1}{z_2} = frac{3-bi}{1-2i} = frac{(3-bi)(1+2i)}{(1-2i)(1+2i)} = frac{3+2b}{5} + frac{6-b}{5}i is a real number, so frac{6-b}{5}=0, solving this gives b=6. Therefore, the answer is: boxed{A}. By using the operation rules of complex numbers and the necessary and sufficient condition for a complex number to be real, we can obtain the answer. This question tests the operation rules of complex numbers, the necessary and sufficient condition for a complex number to be real, and examines reasoning and computational skills, making it a basic question.
question:A traffic light has the following cycle: green for 50 seconds, yellow for 5 seconds, and then red for 40 seconds. Leah picks a random fivesecond time interval to watch the light. What is the probability that the color changes while she is watching?
answer:The light completes a full cycle every 50 + 5 + 40 = 95 seconds. Leah witnesses the color change if she begins to look within five seconds before the transition from green to yellow, yellow to red, or red to green (before the next cycle starts). - The green to yellow transition can start anytime within the last 5 seconds of the 50 seconds allotted to green. - The yellow to red transition can begin anytime in the last 5 seconds of the 5 seconds allotted to yellow. - The red to green transition (and thus the start of the next cycle) can start anytime in the last 5 seconds of the 40 seconds allotted to red. Summing up these intervals: 5 + 5 + 5 = 15 potential seconds during which Leah could start watching and see a color change. The probability that Leah witnesses a color change is therefore: frac{15}{95} = frac{3}{19} Conclusion: The probability that Leah witnesses a color change during her observational interval is boxed{frac{3}{19}}.
question:f(x) is an even function on mathbb{R}, and when x geq 0, f(x) = x^3 + ln(x+1). Then, when x < 0, f(x) = （ ） A: x^3 ln(x1) B: x^3 + ln(x1) C: x^3 ln(1x) D: x^3 + ln(1x)
answer:Since f(x) is an even function on mathbb{R}, we have f(-x) = f(x). When x geq 0, f(x) = x^3 + ln(x+1), then, when x < 0, f(x) = f(-x) = -x^3 + ln(1-x). Therefore, the correct option is: boxed{text{D}}. This problem can be solved by utilizing the property of even functions and transforming the given conditions accordingly. It examines the application of the evenness of functions and the method of finding the expression of a function, making it a basic question.
question:Karen wanted to go out to get some fast food. She pulls up to the drivethrough and orders a 5dollar burger. Her son then yelled out that he wanted a sandwich, so it was added to the order. Karen then decided to order some drinks and opted for two 4dollar smoothies. The total cost of Karen's fastfood order is 17. How much does the sandwich cost?
answer:Let's break down the cost of Karen's order: - The burger costs 5. - The two smoothies cost 2 x 4 = 8. So the cost of the burger and the smoothies together is 5 + 8 = 13. The total cost of the order is 17. To find the cost of the sandwich, we subtract the cost of the burger and the smoothies from the total cost: 17 (total cost) - 13 (cost of burger and smoothies) = 4. Therefore, the sandwich costs boxed{4} .