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question:Daniel practices basketball for 15 minutes each day during the school week. He practices twice as long each day on the weekend. How many minutes does he practice during a whole week?
answer:To calculate the total minutes Daniel practices basketball during a whole week, we break it down into two parts: the school week and the weekend. 1. **During the School Week:** - Daniel practices for 15 minutes each day. - There are 5 school days in a week. - Therefore, the total practice time during the school week is 15 times 5 = 75 minutes. 2. **During the Weekend:** - Daniel practices twice as long each day on the weekend compared to a school day. - The length of practice each day on the weekend is 15 times 2 = 30 minutes. - There are 2 days in a weekend. - Therefore, the total practice time during the weekend is 30 times 2 = 60 minutes. 3. **Total Practice Time in a Week:** - To find the total practice time for the whole week, we add the practice times for the school week and the weekend. - The total practice time is 75 + 60 = 135 minutes. Thus, Daniel practices for a total of boxed{135} minutes during a whole week.
question:triangle ABC and triangle A'B'C', angle B = angle B' = text{Rt}angle, angle A = 30^circ, then among the following conditions, the one that cannot prove triangle ABC is similar to triangle A'B'C' is ( ) A: angle A' = 30^circ B: angle C = 60^circ C: angle C' = 60^circ D: angle A' = 2angle C'
answer:**Solution**: A, Since angle A' = 30^circ, angle B = angle B' = 90^circ, angle A = 30^circ, Therefore, triangle ABC sim triangle A'B'C', so this option is incorrect; B, Given angle C = 60^circ, it's impossible to determine the angles in triangle A'B'C', hence it cannot be proven that triangle ABC sim triangle A'B'C', Therefore, this option is correct; C, Since angle C' = 60^circ, then angle A' = 30^circ, Since angle B = angle B' = 90^circ, angle A = 30^circ, therefore, triangle ABC sim triangle A'B'C', so this option is incorrect; D, Since angle A' = 2angle C', angle A' + angle C' = 90^circ, Therefore, angle A' = 30^circ, Since angle B = angle B' = 90^circ, angle A = 30^circ, Therefore, triangle ABC sim triangle A'B'C', so this option is incorrect. Hence, the correct choice is boxed{B}.
question:Calculate the sum of the following series of fractions: dfrac{1}{12}+dfrac{2}{12}+dfrac{3}{12}+dfrac{4}{12}+dfrac{5}{12}+dfrac{6}{12}+dfrac{7}{12}+dfrac{8}{12}+dfrac{9}{12}+dfrac{12}{12}+ dfrac{15}{12}+dfrac{18}{12} text{(A)} 6 qquad text{(B)} 7.5 qquad text{(C)} 8 qquad text{(D)} 9 qquad text{(E)} 10
answer:1. **Convert fractions to a common denominator**: All fractions have the same denominator, 12, so sum the numerators directly. 2. **Add the numerators**: The sequence of numerators is 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 18. The sum of these numbers: [ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 12 + 15 + 18 = 90 ] 3. **Divide the total sum by the common denominator**: Since the total of numerators is 90 and the common denominator is 12: [ frac{90}{12} = 7.5 ] 4. **Conclusion**: The sum of the series of fractions is 7.5. [ 7.5 ] The final answer is boxed{text{(B)} 7.5}
question:f 150 is divided into three parts proportional to 3, 5, and frac{7}{2}, the smallest part is? textbf{(A)} frac{900}{23}qquadtextbf{(B)} 50qquadtextbf{(C)} 30qquadtextbf{(D)} 20qquadtextbf{(E)} text{None of these answers}
answer:1. **Identify Proportions**: The parts are in proportion to 3, 5, and frac{7}{2}. This set cannot be simplified to integers due to the fraction frac{7}{2}. 2. **Set Up Variable Ratio**: Let the three parts be 3k, 5k, and frac{7k}{2} respectively, where k is the common variable factor in the ratio. 3. **Sum Equation Formulation**: The sum of these parts must equal 150. Therefore, the equation is: [ 3k + 5k + frac{7k}{2} = 150 ] Simplifying, we combine terms: [ 8k + frac{7k}{2} = 150 ] Convert 8k to a fraction to unify the equation: [ frac{16k}{2} + frac{7k}{2} = 150 ] [ frac{23k}{2} = 150 ] 4. **Solve for k**: [ 23k = 300 quad (text{by multiplying both sides by } 2) ] [ k = frac{300}{23} ] 5. **Find Smallest Part**: The smallest part corresponds to 3k: [ 3k = 3 cdot frac{300}{23} = frac{900}{23} ] Finally, the smallest part is: [ frac{900{23}} ] The final answer is boxed{textbf{(A)} frac{900}{23}}