Three 19, ∫≈ AEC = ∠ ADC = ∠ B = 90, ∴OE=OD=OB (the median line on the hypotenuse of a right triangle is equal to half of the hypotenuse), ∴E, D, BE⊥DE are all in the diameter of AC.
20. AB=AC, triangle ABP and ACQ are congruent, AQ=AP, angle BAP= angle CAQ, angle PAQ= angle BAC=60 degrees, triangle APQ is isosceles, and one angle is 60 degrees. Triangle APQ is an equilateral triangle 2 1, and ADC area: area of triangle ABC = 2: 3, so AD: BC = 2: 3. Because MN= 10 prolongs the intersection of MN and NM in E and F respectively, MF = 1/2ad, NE= 1/2AD, EF = 1/2.
22. Test site: triangle midline theorem.
Title: Proof. Analysis: Extend the intersection of AD and BC in F, and prove that AC=CF and DE are the midline of △ABF, which can be verified. Solution: Extending the intersection of AD and BC in F means that AC=CF and DE is the center line of △ABF.
* CD sharing ∠ACB, AD⊥CD,
∴∠ACD=∠BCD, CD is the public side, ∠ ADC =∠ FDC = 90,
∴AC=CF,AD=FD
E in ∵△ABC is the midpoint of AB,
∴DE is the center line of △ABF,
∴de= 12bf= 12(bc-cf)= 12(bc-ac).
23. it is proved that the intersection o is og∨BC, AE is in g∫ square ABCD ∴ AO = Co, ∠Abd =∠ACB = 45∫og∨BC∴aog =∠ACG.
24. (1) If the number of people who donated 15 yuan is 5x, then the number of people who donated 20 yuan is 8x.
∴5x+8x=39,∴x=3
∴ A * * * Survey 3x+4x+5x+8x+2x=66 (person) (3 points).
∴ The probability that the donation is not less than 20 yuan is 30 66 = 511. (5 points)
(2) According to (1), the mode of this set of data is 20 yuan, and the median is 15 yuan. (7 points)
(3) Donations from all students:
(9× 5+12×10+15×15+24× 20+6× 30) ÷ 66× 2310 = 36750 (.
25. At the beginning of (1), the wind speed increased by 2 km/h on average, and after 4 hours, the wind speed was 2 * 4 = 8 km/h .. After 6 hours (to the first 10 hour), the wind speed increased by 4 km/h on average, so the wind speed was 8+4*6=32 km/h ... (2) From 10-25, the wind speed of sandstorm will be maintained at 32km per hour. After encountering the green vegetation area, its wind speed drops by 1 km/h on average every hour, and finally stops. So the sandstorm will stop after 32/ 1=32 hours. So from the sandstorm to the end, ***25+32=57 hours. (3) According to the image, the CD passes through (25, 32)(57, 0).
Let the resolution function be y = kx+b.
∴{ 25k+b = 3257k+b = 0 { k =- 1b = 57。
∴y=-x+57(25≤x≤57)；
26. Solution: (1) X tons of cement is set for warehouse A. According to the meaning of the question, y =12× 20x+10× 25 (100-x)+12×/kloc. The value of k =-30 < 0 ∴ y in the above linear function decreases with the increase of x: when ∴ x = 70, the total freight (y) is the least, and the minimum total freight is 37 100 yuan. Winter vacation homework 2
(3) 15, and the math score can be set as x (85+96+x)/3 > = 90 85+96+x > = 270 X>= 89 The math score is not less than 89 points.
16 is sold at a discount of x from the original price, so:1000×1/2×10+1000×1/2×10× = 7× 1000+20005000+500 x & gt； =90005x >= 40x & gt=8, so the maximum discount is 20%.
17， 10x+x-3 & lt； 52 x
20, x+2y=6, x-y=9-3a, and the solutions x and y are both positive numbers. First, x = 8-2a, y =-1+ax = 8-2a >; 0 and y =-1+a >; 0a<4 and a> 1 means 1
22, suppose that each group initially produces x parts per day on average (x is a positive integer) 10x.
23. (1)1000 x0.1=100, less than 200. (2) 140- 100 = 40, 40 divided by 0.2 = 200,65438. Divided by 0.2=500,1000+500 = 15001200 to1500.
24. (1) If you plan to buy X seedlings of the first category, you plan to buy 500-x seedlings of the second category: 50x+80 (500-x) = 2800050x+40000-80x = 28000-30x =-120. Answer: Buy 400 seedlings A and 0/00 seedlings B/KLOC (2) 50x+80 (500-X) ≤ 3400050x+40000-80x ≤ 34000-30x ≤-6000x ≥ 200 plants (pieces); Buy no less than 200 A-type seedlings, and choose B-type seedlings for the rest; (3) If you plan to buy X seedlings of A, you plan to buy 500-x seedlings of the second kind: 90% x+95% (500-x) ≥ 92% x 500x ≤ 300 500-x ≥ 20050x+80 (500-x) ≤ 34 000x ≥ 200. =-30x+40000 are all solutions. I searched one question after another and sorted it out, but I couldn't find two. Pay attention to the question number. I don't want to find another class. Think for yourself. Ha ha. . . . . .