The ratio of extracurricular books between Xiaohong and Xiaoli is 2:1. If Xiaohong lends 11 books to Xiaoli, the ratio of extracurricular books between Xiaohong and Xiaoli is 7:20 How many extra-curricular books did Xiao Hong and Xiao Li have?

The ratio of extracurricular books between Xiaohong and Xiaoli is 2:1. If Xiaohong lends 11 books to Xiaoli, the ratio of extracurricular books between Xiaohong and Xiaoli is 7:20 How many extra-curricular books did Xiao Hong and Xiao Li have?

Let's suppose that Xiaohong has 2x extra-curricular books and Xiaoli has x extra-curricular books
(2x-11):(x+11)=7:20
7(x+11)=20(2x-11)
7x+77=40x-220
40x-7x=220+77
33x=297
X=9
2 * 9 = 18
A: it turns out that Xiaohong has 18 extra-curricular books and Xiaoli has 9 extra-curricular books
Reading the same book, Xiao Hong read 14 pages in 3 days, Xiao Li read 23 pages in 6 days, which of them read fast?
6 divided by 3 is 2
2 times 14 equals 28
28 ＞ 23
Xiaohong Xiaoli
A: Xiaohong is quick to see
Xiaohong and Xiaoli use the temperature difference to measure the height of the mountain. Xiaohong's temperature at the top of the mountain is - 1 ℃, and Xiaoli's temperature at the foot of the mountain is 5 ℃
The temperature rises by 0.6 degrees per 100 meters
The total drop is 6 degrees, that is 100 * (6 / 0.6) = 1000m
Xiaohong and Xiaoli use the temperature difference to measure the height of a mountain. Xiaohong's temperature at the top of the mountain is - 1 degrees, while Xiaoli's temperature at the foot of the mountain is 5 degrees
The temperature is about 0.6 degrees lower in 100 meters. How many meters is the height of the mountain?
﹙－1－5﹚÷﹙－0.6﹚×100=1000
Xiaohong and Xiaoli use the temperature difference to measure the height of the mountain peak. Xiaohong measured the temperature at the mountain peak at 1.25 degrees, while Xiaoli measured the temperature at the foot of the mountain at 6.28 degrees. It is known that every 100 meters increase in the height of the area, the temperature will decrease by about 0.8 degrees
No equations
6.28-1.25=5.03
5.03/0.8=6.2875
6.2875 * 100 = 628.75m
Electric work and energy
Electric power index_________ The essence of the work done by the current is to________ Can be converted into_____ In the process of energy conversion, how much work is done by the current, how much electric energy is converted into other forms of energy. Both electric energy and electric work are represented by symbols, and the unit is____ . calculation formula w=______
Electric power index___ Current___ The essence of the work done by the current is to___ Electricity___ Can be converted into___ Other forms__ In the process of energy conversion, how much work is done by the current, how much electric energy is converted into other forms of energy. Both electric energy and electric work are represented by symbols, and the unit is__ J__ . calculation formula w=___ Pt=IUt___
Electric power index__ Current_______ The essence of the work done by the current is to____ Electricity____ Can be converted into__ Other forms___ In the process of energy conversion, how much work is done by the current, how much electric energy is converted into other forms of energy. Both electric energy and electric work are represented by symbols in units of__ Joule. Calculation formula w=__ I^2*R*t____
Mathematics in Grade 6 of primary school oral arithmetic
1 / 7 and 5 / 9 times 7 =?
211 divided by 7 minus 1 / 7 =?
8 / 9 times (1.5 times 3 / 4) =?
=5 out of 9
=30
=9 out of 8 = 1 and 1 out of 8
Let the sum of the first n terms of the arithmetic sequence {an} be Sn, and A3 = 5, S15 = 225. (I) find the general term an of the sequence {an}; (II) let BN = 2An + 2n, find the sum of the first n terms and TN of the sequence {BN}
(I) let the first term of the arithmetic sequence {an} be A1, and the tolerance be d. from the meaning of the problem, we get a1 + 2D = 515a1 + 15 × 142d = 225, and the solution is A1 = 1D = 2, an = 2N-1; (II) BN = 2An + 2n = 12.4n + 2n, TN = B1 + B2 + +bn=12（4+42+… +4n）+2（1+2+… +n）=4n+1−46+n2+n=23•4n+n2+n−23．
When the temperature drops to a very low value, the resistance of some metal conductors will suddenly disappear. This phenomenon is called superconductivity, and this temperature is called transition temperature. For example, when the temperature drops to - 269 ℃, the resistance of mercury disappears. Superconductivity is one of the major discoveries in the 20th century. Many scientists all over the world are devoted to the study of superconductors, The research on superconductors in China has been in the forefront of the world
What is superconductivity?
Can the resistance wire of electric furnace be made of superconducting material? (assuming the temperature is low enough) (hint: the heat of electric furnace wire is q = i2rt)
If the transmission line is made of superconductive material, what are the advantages and disadvantages?
1. When the temperature drops to a very low value, the resistance of some metal conductors will suddenly disappear. This phenomenon is called superconductivity. 2. No, the temperature is still high. The electric furnace wire must not be made of superconductive material, otherwise the electric furnace wire will not emit any heat
Ask for 100 oral arithmetic questions in Grade 6
To primary school sixth grade oral problems 100, do not use the problem! As long as the oral fast!
2.8×0.4＝ 1.12
14－7.4＝6.6,
1.92÷0.04＝48,
0.32×500＝160,
0.65＋4.35＝ 5
10－5.4＝4.6,
4÷20＝0.2,
3.5×200＝700,
1.5－0.06＝1.44
0.75÷15＝0.05,
0.4×0.8＝0.32,
4×0.25＝1,
0.36＋1.54＝2
1.01×99＝99.99,
420÷35＝12,
25×12＝300,
135÷0.5＝270
3/4 + 1/4 =1,
2 + 4/9 =22/9,
3 - 2/3 =7/3,
3/4 - 1/2= 1/4
1/6 + 1/2 -1/6 =1/2,
7.5-(2.5+3.8)=1.2,
7/8 + 3/8 =5/4
3/10 +1/5 =1/2,
4/5 - 7/10 =1/10,
2 - 1/6 -1/3 =1.5
0.51÷17＝0.03,
32.8＋19＝51.8,
5.2÷1.3＝4,
1.6×0.4＝ 0.64
4.9×0.7＝3.43,
1÷5＝0.2,
6÷12＝0.5,
0.87－0.49＝0.38
1.(1+1/2)(1+1/3)(1+1/4).(1+1/100)
2.(1-1/2)(1-1/3)(1-1/4).(1-1/100)
3.8+2-8+2
4.25*4/25*4
5.7.26-(5.26-1.5)
6.286+198
7.314-202
8.526+301
9.223-99
10.6.25+3.85-2.125+3.875
11.9-2456*21
12.0.5/11.5-4*2.75
13.1/2×3/5
14.3.375+5.75+2.25+6.625
15.1001－9036÷18
16.3.8×5.25+14.5
17.2.1*4.3+5.7*2.1
18.30×1/3
19.102*45-328
20.2/3×12
21.2.8*3.1+17.6/8
22.3/5×5/6
23.(50-12.5)/2.5
24.2/5×1/3
25.6110*47+639
26.1/2-1/6
27.3.5*2.7-52.2/18
28.1/7×1/5
29.3.375*0.97+0.97*6.625
30.25×4/5
31.6.54+2.4+3.46+0.6
32.5/6-1/2
33.95.6*1.8+95.6*8.2
34.1/2×1/5
35.600-420/12
36.344/3.6-5.4*0.25
37.16/2+30/2+90/6
38.3001-1998.
39.5000-105*34
40.0.15/0.25+0.75*1.2
41.(1/2+1/3+1/4)*0.24
42.(25+4)*4
43.300-4263/21
44.0.81/0.25+5.96
45.403÷13×27
46.1.5×4.2－0.75÷0.25
47.3.27×4 +3.27×5.7
48.（1.2+ 1.8）×4.51025－768÷32
49.0.25×80－0.45÷0.9
50.1025－768÷32
51.0.25*2.69*4
52.2348+275*16
53.2/9*15/8-1/12*9/5
54.2.4+2.4*(5.375-3.375)
55.645-45*12
56.0.15+1.2/0.24-0.45
57.3.75-(2.35+0.25/1.25)
58.76*1/4+23*25/100+0.25
59.10-2.87-7.13
60.0.96+9.6*9.9
61.7.5-5.7*1/3
62.12.37-3.25-6.75
63.16*6.8+2.2*16+16
64.401*19+284
65.58.7-16.65/3.7
66.0.4*4.7*2.5+(2.3+5.3)
67.9.31-1.125-7.875
68.640+128*45
69.8.2*1.6-0.336/4.2
70.400*(0.62+0.08)
2/1*2=1
3/1*3=1
3/2*3=2
3/1*6=2
4/3*8=6
5/3*20=12
7/3*14=6
8/7*40=35
4/3*16=12
9/5*27=15
2/1*30=15
12/7*24=14
30/1*30=1
51/9*102=18
19/9*76=36
4/9*8=18
5/8*90=144
99/98*99=98
3/14*6=28
7/1*28=4
10/1*90=9
5/3*105=63
19/7*38=14
5/1*25=5
8/19*16=38
61/60*122=120
7/2*28=8 6/1*48=8
9/7*18=14
25/7*100=28
9/5*81=45 8/9*16=18
2/1*2=1
3/1*3=1
3/2*3=2
3/1*6=2
4/3*8=6
5/3*20=12
7/3*14=6
8/7*40=35
4/3*16=12
9/5*27=15
2/1*30=15
12/7*24=14
30/1*30=1
51/9*102=18
19/9*76=36
4/9*8=18
5/8*90=144
99/98*99=98
3/14*6=28
7/1*28=4
10/1*90=9
5/3*105=63
19/7*38=14
5/1*25=5
8/19*16=38
61/60*122=120
7/2*28=8
6/1*48=8
9/7*18=14
25/7*100=28
9/5*81=45
8/9*16=18
12÷3/5=12×(5/3）
9÷6/7=9×( 7/6 )
30÷5/6=30×（6/5 ）
4×（3/2 ）=4÷2/3
4÷5/7=4×7/5
3÷4/5=3×5/4
24÷7/16=24×（16/7 ）
A÷C/B=A×B/C
4÷4/5=5
6÷3/4=8
10÷2/5=25
18÷4/9=81/2
4×4/5=16/5
6×3/4=18/4
10×2/5=4
18×4/9=8
3÷3/4=4
2÷1/3=6
6÷4/5=15/2
1÷5/7=7/5
3/4÷3=1/4
1/3÷2=1/6
4/5÷6=2/15
5/7÷1=5/7
2/1*2=1
3/1*3=1
3/2*3=2
3/1*6=2
4/3*8=6
5/3*20=12
7/3*14=6
8/7*40=35
4/3*16=12
9/5*27=15
2/1*30=15
12/7*24=14
30/1*30=1
51/9*102=18
19/9*76=36
4/9*8=18
5/8*90=144
99/98*99=98
3/14*6=28
7/1*28=4
10/1*90=9
5/3*105=63
19/7*38=14
5/1*25=5
8/19*16=38
61/60*122=120
7/2*28=8
6/1*48=8
9/7*18=14
25/7*100=28
9/5*81=45
8/9*16=18
2/1*2=1
3/1*3=1
3/2*3=2
3/1*6=2
4/3*8=6
5/3*20=12
7/3*14=6
8/7*40=35
4/3*16=12
9/5*27=15
2/1*30=15
12/7*24=14
30/1*30=1
51/9*102=18