Appearance
question:particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 7 % . what was its profit on the items sold in february ?
answer:Let's assume the cost price of the turtleneck sweaters is ( C ). The store first marked up its cost by 20%, so the initial selling price (SP1) would be: [ SP1 = C + 0.20C = 1.20C ] During the New Year season, it marked up its prices by 25% of the original retail price (which is SP1), so the second selling price (SP2) would be: [ SP2 = SP1 + 0.25SP1 = 1.25SP1 = 1.25(1.20C) = 1.50C ] In February, the store offered a discount of 7% on SP2. The final selling price (SPF) after the discount would be: [ SPF = SP2 - 0.07SP2 = 0.93SP2 = 0.93(1.50C) = 1.395C ] The profit on the items sold in February would be the final selling price minus the cost price: [ Profit = SPF - C = 1.395C - C = 0.395C ] So the store's profit on the items sold in February is boxed{39.5%} of the cost price.
question:Given the function f(x)=cos ^{2}( frac {π}{4}+x)cos ^{2}( frac {π}{4}x), find the value of f( frac {π}{12}). The options are: A: frac {1}{2} B: frac {1}{2} C: frac { sqrt {3}}{2} D: frac { sqrt {3}}{2}
answer:The function can be simplified as follows: f(x)=cos ^{2}( frac {π}{4}+x)-cos ^{2}( frac {π}{4}-x) Using the double angle identity for cosine, we can rewrite cos ^{2}( frac {π}{4}+x) as frac{1}{2}(1 + cos(2(frac {π}{4}+x))). Similarly, using the double angle identity for sine, we can rewrite cos ^{2}( frac {π}{4}-x) as frac{1}{2}(1 - sin(2(frac {π}{4}-x))). Substituting these expressions into the original equation, we get: f(x)=frac{1}{2}(1 + cos(2(frac {π}{4}+x))) - frac{1}{2}(1 - sin(2(frac {π}{4}-x))) Simplifying further, we get: f(x)=frac{1}{2}(cos(2(frac {π}{4}+x)) + sin(2(frac {π}{4}-x))) Using the sum and difference formulas for cosine and sine, we can simplify the expression to: f(x)=frac{1}{2}(cos(frac {π}{2}+2x) + sin(frac {π}{2}-2x)) Using the cofunction identities, we can simplify the expression to: f(x)=frac{1}{2}(-sin(2x) + cos(2x)) Using the double angle identity for sine, we can simplify the expression to: f(x)=-sin(2x) Substituting x = frac {π}{12}, we get: f(frac {π}{12})=-sin(frac {π}{6}) Since sin(frac {π}{6}) = frac{1}{2}, we have: f(frac {π}{12})=-frac{1}{2} Therefore, the answer is boxed{-frac{1}{2}}.
question:A train of a certain length can cross an electric pole in 3 sec with a speed of 179.99999999999997. What is the length of the train?
answer:To find the length of the train, we can use the formula: Length of the train = Speed of the train x Time taken to cross the electric pole Given that the speed of the train is 179.99999999999997 m/s (which we can round to 180 m/s for simplicity) and the time taken to cross the electric pole is 3 seconds, we can calculate the length as follows: Length of the train = 180 m/s * 3 s = 540 meters Therefore, the length of the train is boxed{540} meters.
question:John is very unfit and decides to work up to doing a pushup. He trains 5 days a week for them and starts with wall pushups. He adds 1 rep a day and once he gets to 20 reps, he will progress to the next pushup variation. The variations he will train in order are wall pushups, incline pushups, knee pushups, and then floor pushups. Each time he reaches 20 reps, he will move on to the next variation. How many weeks will it take him to reach floor pushups?
answer:Let's break down the progression: 1. Wall push-ups: John starts with 1 rep and adds 1 rep each day. He needs to reach 20 reps to move to the next variation. This will take him 20 days. 2. Incline push-ups: After 20 days of wall push-ups, he will start incline push-ups at 1 rep and again add 1 rep each day until he reaches 20 reps. This will take another 20 days. 3. Knee push-ups: After 20 days of incline push-ups, he will start knee push-ups at 1 rep and add 1 rep each day until he reaches 20 reps. This will take another 20 days. So, in total, John will take 20 (wall) + 20 (incline) + 20 (knee) = 60 days to reach floor push-ups. Since John trains 5 days a week, we need to calculate how many weeks this will take. 60 days / 5 days per week = 12 weeks Therefore, it will take John boxed{12} weeks to progress to floor push-ups.