Appearance
question:A certain machine is capable of producing eight rulers per minute. How many rulers can this machine produce in 15 minutes? (a) 104 (b) 110 (c) 112 (d) 128 (e) 120
answer:1. **Understand the given data**: The machine produces 8 rulers per minute. 2. **Determine the total time**: We need the number of rulers produced in 15 minutes. 3. **Calculate the total number of rulers**: [ text{Total rulers} = text{rulers per minute} times text{total time} ] Substituting the given values: [ text{Total rulers} = 8 , text{rulers per minute} times 15 , text{minutes} = 8 times 15 ] 4. **Perform the multiplication**: [ 8 times 15 = 120 ] 5. **Conclusion**: The machine can produce 120 rulers in 15 minutes. Thus, the correct answer is [ boxed{120} ]
question:A certain commodity has a cost price of a yuan per unit. It is sold at a markup of 25% on the cost price. Due to overstocking, the price is reduced to 90% of the selling price. At this point, the selling price is ( ) A: 1.125a yuan B: 1.25a yuan C: 0.75a yuan D: 1.5a yu
answer:To solve this problem, we start by understanding that the commodity is initially marked up by 25% on its cost price, a yuan. This markup can be mathematically represented as an increase of 25% on a, which is equivalent to multiplying a by 1 + 25% or 1.25. This gives us the selling price before any discount is applied. Next, due to overstocking, the price is reduced to 90% of this new selling price. Mathematically, this reduction can be represented by multiplying the selling price by 0.9. Putting these two steps together, we can calculate the final selling price as follows: begin{align*} text{Final Selling Price} &= aleft(1+25%right)times 0.9 &= a times 1.25 times 0.9 &= 1.125a text{ yuan} end{align*} Therefore, the final selling price of the commodity, after the markup and the subsequent reduction, is 1.125a yuan. This matches option A in the multiple-choice answers. Thus, the correct answer is boxed{A}.
question:A man has Rs. 10350 in the form of Rs. 50 notes and Rs. 500 notes. He has 77 notes of Rs. 50 denomination. What is the total number of notes he has?
answer:Let's denote the number of Rs. 500 notes as ( x ). The total amount of money from the Rs. 50 notes is ( 77 times 50 ). The total amount of money from the Rs. 500 notes is ( x times 500 ). According to the information given, the sum of these amounts is Rs. 10350. So we can write the equation: ( 77 times 50 + x times 500 = 10350 ) Now, let's solve for ( x ): ( 3850 + 500x = 10350 ) Subtract 3850 from both sides: ( 500x = 10350 - 3850 ) ( 500x = 6500 ) Now, divide both sides by 500 to find ( x ): ( x = frac{6500}{500} ) ( x = 13 ) So, the man has 13 notes of Rs. 500 denomination. To find the total number of notes, we add the number of Rs. 50 notes to the number of Rs. 500 notes: Total number of notes = Number of Rs. 50 notes + Number of Rs. 500 notes Total number of notes = 77 + 13 Total number of notes = 90 Therefore, the man has a total of boxed{90} notes.
question:Given f(x)=2cos(xfrac{π}{2})+f'(0)cos x, then the slope of the tangent line to the curve y=f(x) at the point ({frac{{3π}}{4}},f({frac{{3π}}{4}})) is () A: sqrt{2} B: sqrt{2} C: 2sqrt{2} D: 2sqrt{2}
answer:To find the slope of the tangent line to the curve y=f(x) at the point left(frac{3pi}{4}, fleft(frac{3pi}{4}right)right), we start by understanding the given function and its derivative. Given function: [f(x) = 2cosleft(x - frac{pi}{2}right) + f'(0)cos x] We can use the trigonometric identity cosleft(alpha - frac{pi}{2}right) = sinalpha to rewrite f(x) as: [f(x) = 2sin x + f'(0)cos x] Taking the derivative of f(x) with respect to x, we get: [f'(x) = frac{d}{dx}left(2sin x + f'(0)cos xright)] [f'(x) = 2cos x - f'(0)sin x] Given that f'(0) = 2, we substitute this value into the expression for f'(x): [f'(x) = 2cos x - 2sin x] Now, to find the slope of the tangent line at x = frac{3pi}{4}, we evaluate f'left(frac{3pi}{4}right): [f'left(frac{3pi}{4}right) = 2cosleft(frac{3pi}{4}right) - 2sinleft(frac{3pi}{4}right)] [= 2left(-frac{sqrt{2}}{2}right) - 2left(frac{sqrt{2}}{2}right)] [= -2sqrt{2}] Therefore, the slope of the tangent line to the curve at the point left(frac{3pi}{4}, fleft(frac{3pi}{4}right)right) is boxed{-2sqrt{2}}, which corresponds to choice D.