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question:triangle ABC, if b = 2a and B = A + 60^circ, then A = ______
answer:Given that in triangle ABC, b = 2a and B = A + 60^circ. According to the law of sines, we have: frac{a}{sin A} = frac{b}{sin B} Substituting b = 2a and B = A + 60^circ into the equation, we get: frac{a}{sin A} = frac{2a}{sin(A + 60^circ)} Simplifying, we find: sin(A + 60^circ) = 2sin A Using the sine addition formula, sin(A + 60^circ) = sin A cos 60^circ + cos A sin 60^circ, we can rewrite the equation as: sin A cdot frac{1}{2} + cos A cdot frac{sqrt{3}}{2} = 2sin A Solving for sin A, we get: frac{sqrt{3}}{2}cos A = frac{3}{2}sin A This implies that tan A = frac{1}{sqrt{3}}, which means A = 30^circ. Therefore, A = boxed{30^circ}.
question:the sequence {a_n}, (a_1=1, a_{n+1}=frac{a_n}{ccdot a_n+1}) (c is a constant, ninmathbb{N^*}), and (a_1,a_2,a_5) form a geometric sequence with a common ratio not equal to 1. (1) Prove that the sequence {frac{1}{a_n}} is an arithmetic sequence and find the value of c; (2) Let b_n=a_n a_{n+1}, find the sum of the first n terms of the sequence {b_n}, denoted as (S_n).
answer:(1) Given a_1=1, a_{n+1}=frac{a_n}{ccdot a_n+1}, it follows that a_nneq 0, then frac{1}{a_{n+1}}-frac{1}{a_n}=frac{ccdot a_n+1}{a_n}-frac{1}{a_n}=c, and c is a constant, therefore the sequence {frac{1}{a_n}} is an arithmetic sequence; We know that frac{1}{a_n}=frac{1}{a_1}+(n-1)c=1+(n-1)c, since a_1=1, therefore a_2=frac{1}{1+c}, a_5=frac{1}{1+4c}, given that a_1, a_2, a_5 form a geometric sequence with a common ratio not equal to 1, hence (frac{1}{1+c})^2=frac{1}{1+4c}, solve for c=0 or c=2, when c=0, a_n=a_{n+1}, which does not meet the requirement, discard this solution, therefore the value of c is 2; (2) from (1) we know c=2, therefore a_n=frac{1}{2n-1}, b_n=a_n a_{n+1}=frac{1}{2n-1}cdotfrac{1}{2n+1}=frac{1}{2}(frac{1}{2n-1}-frac{1}{2n+1}), thus, the sum of the first n terms of the sequence {b_n} S_n=frac{1}{2}[(1-frac{1}{3})+(frac{1}{3}-frac{1}{5})+ldots+(frac{1}{2n-1}-frac{1}{2n+1})]=boxed{frac{1}{2}(1-frac{1}{2n+1})}
question:Which of the following statements is (are) equivalent to the statement If a car is red, then it is not a sports car? textbf{(A)} textbf{I.} text{and} textbf{II.} text{only} qquad textbf{(B)} textbf{III.} text{and} textbf{IV.} text{only} qquad textbf{(C)} textbf{I.} text{and} textbf{IV.} text{only} qquad textbf{(D)} textbf{I.} text{and} textbf{III.} text{only} qquad textbf{(E)} textbf{III.} text{only}
answer:Let's denote the statement "a car is red" as ( P ) and "it is not a sports car" as ( Q ). The original statement can then be written as: [ P implies Q ] We need to analyze the logical equivalence of the following statements: - **I.** ( P implies Q ) - **II.** ( Q implies P ) - **III.** ( neg Q implies neg P ) - **IV.** ( neg P lor Q ) Step 1: Analyze Statement I Statement I is the given statement: [ P implies Q ] This is the base statement and is obviously equivalent to itself. Step 2: Analyze Statement II Statement II is: [ Q implies P ] This is the converse of the original statement. The converse of a statement is not necessarily logically equivalent to the original statement unless the original statement is a biconditional (if and only if). Therefore, ( Q implies P ) is not equivalent to ( P implies Q ) unless additional information is given. Step 3: Analyze Statement III Statement III is: [ neg Q implies neg P ] This is the contrapositive of the original statement. By logical equivalence, the contrapositive of a statement is always true if the original statement is true. Therefore, ( neg Q implies neg P ) is equivalent to ( P implies Q ). Step 4: Analyze Statement IV Statement IV is: [ neg P lor Q ] This is the disjunction form equivalent to the original implication. The statement ( neg P lor Q ) is equivalent to ( P implies Q ) by the implication rule, where ( P implies Q ) is logically equivalent to ( neg P lor Q ). # Conclusion: From the analysis, Statements I, III, and IV are equivalent to the original statement ( P implies Q ). Therefore, the correct answer is textbf{III. text{and} textbf{IV.} text{only}}. The final answer is boxed{textbf{(B)} textbf{III.} text{and} textbf{IV.} text{only}}
question:Keanu was surfing towards the shore at a speed of 20 miles per hour. A shark was swimming alongside him, riding the same wave, when it doubled its speed and moved away from Keanu. There was a pilot fish swimming alongside Keanu and the shark, and when the shark increased its speed, the pilot fish also increased its speed and moved away from Keanu, but only increasing its speed by half as much as the shark had increased its speed by. What speed, in miles per hour, was the pilot fish swimming when it moved away from Keanu?
answer:To solve this problem, we follow the given information step by step: 1. The shark initially matches Keanu's speed of 20 miles per hour. When it doubles its speed, we calculate the new speed of the shark as: [ 20 , text{mph} times 2 = 40 , text{mph} ] 2. The increase in the shark's speed can be found by subtracting its initial speed from its new speed: [ 40 , text{mph} - 20 , text{mph} = 20 , text{mph} ] 3. The pilot fish increases its speed by half the amount that the shark increased its speed by. Therefore, the increase in the pilot fish's speed is: [ frac{20 , text{mph}}{2} = 10 , text{mph} ] 4. Finally, to find the speed at which the pilot fish moved away from Keanu, we add this increase to Keanu's speed: [ 20 , text{mph} + 10 , text{mph} = 30 , text{mph} ] Therefore, the speed at which the pilot fish was swimming when it moved away from Keanu is boxed{30 , text{mph}}.