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question:There are 10 numbers from a certain number to 65 that are exactly divisible by 5. What is the starting number of this range?
answer:If the numbers are exactly divisible by 5, they must be multiples of 5. Since there are 10 numbers in this range, we can find the starting number by counting backwards from 65 in multiples of 5, ten times. Let's count backwards in multiples of 5: 1. 65 - 5 = 60 2. 60 - 5 = 55 3. 55 - 5 = 50 4. 50 - 5 = 45 5. 45 - 5 = 40 6. 40 - 5 = 35 7. 35 - 5 = 30 8. 30 - 5 = 25 9. 25 - 5 = 20 10. 20 - 5 = 15 So, the starting number of this range is boxed{15} .
question:the complex plane, the point corresponding to the complex number frac{1+i}{i} is located in ( ) A: The first quadrant B: The second quadrant C: The third quadrant D: The fourth quadr
answer:**Analysis** This question tests the algebraic operations of complex numbers and the geometric meaning of complex numbers. According to the given form of the complex number, first perform the division operation of complex numbers, multiply both the numerator and the denominator by the conjugate of the denominator, simplify it to the simplest form, write out the coordinates of the corresponding point in the complex plane, and determine the location of the point. **Solution** Given that the complex number z= frac{-1+i}{i}= frac{(-1+i)(-i)}{i(-i)}=1+i, Therefore, the coordinates of the point corresponding to the complex number are (1,1), Therefore, the corresponding point is in the first quadrant, Hence, the correct answer is boxed{text{A}}.
question:e tap, working alone at its maximum constant rate, can fill a certain sink in 287 seconds. A second tap, working alone at its maximum constant rate, can fill the same sink in 283 seconds. Added to this, a third tap can fill the sink in 325 seconds at its maximum constant rate. If all three taps work together at their respective maximum constant rates, estimate the time it will take to fill the sink.
answer:To solve this problem, we can use the concept of rates. If the first tap can fill the sink in 287 seconds, its rate is 1/287 of the sink per second. Similarly, the second tap's rate is 1/283 of the sink per second, and the third tap's rate is 1/325 of the sink per second. When all three taps are working together, their rates add up. So the combined rate is: 1/287 + 1/283 + 1/325 To add these fractions, we need a common denominator. However, for an estimate, we can use an approximation by noticing that the denominators are all close to 300. So, as an estimate, we can say that each tap fills slightly more than 1/300 of the sink per second. Let's use 300 as a common denominator for an estimate: 1/287 ≈ 1/300 1/283 ≈ 1/300 1/325 ≈ 1/300 So the combined rate is approximately: 1/300 + 1/300 + 1/300 = 3/300 This simplifies to 1/100, meaning the three taps together fill about 1/100 of the sink per second. To find the time it takes to fill the sink, we take the reciprocal of the combined rate: Time = 1 / (1/100) = 100 seconds So, it is estimated that all three taps working together will fill the sink in about boxed{100} seconds. Keep in mind that this is an approximation, and the actual time may be slightly different due to the rounding of the rates.
question:Circle C has a radius of 5 and diameter overline{AB}. Circle D is internally tangent to circle C at point A. Circle E is internally tangent to circle C, externally tangent to circle D, and tangent to overline{AB}. The radius of circle D is four times the radius of circle E. Find the radius of circle D expressed in the form sqrt{p}q, where p and q are positive integers, and compute p+q.
answer:Let the radius of circle E be r and hence the radius of circle D be 4r. Using the point of tangency at A and the diameter overline{AB}, we can describe the geometric relations: - EF = r, the radius of E, - CE = 5 - r, the distance from the center of C to the tangent point on E, - CF = sqrt{25 - 25r} because CF^2 = CE^2 - EF^2 = (5-r)^2 - r^2 = 25 - 10r. The distance from C to D minus the radius of D is: - CD = CA - AD = 5 - 4r. The distance DF from D to F (tangent point of E to overline{AB}) is: - DF = DC + CF = (5 - 4r) + sqrt{25 - 25r}. Since DE = DF (both are tangent distances from D to E): - DE = 8r. The relation creates the equation: - [(5-4r + sqrt{25-25r})^2 + r^2 = 64r^2] Solving this equation is complex, simplify it down to: - [25 - 40r + 16r^2 + 25 - 25r + r^2 = 64r^2] - [50 - 65r + 17r^2 = 64r^2], - [47r^2 - 65r + 50 = 0]. Solving this using the quadratic formula, we find: - [r = frac{65 pm sqrt{4225 - 9400}}{94}], - [r = frac{65 pm sqrt{5325}}{94}]. Substituting r into 4r we get: - (4r = 4(frac{65 + sqrt{5325}}{94}) = sqrt{5325} - frac{135}{47}), - Thus, p = 5325 and q = 135. Final answer: boxed{5460}.