38400km is represented by 2 significant digits reserved by scientific counting method, and 38400 is

38400km is represented by 2 significant digits reserved by scientific counting method, and 38400 is

38400KM =3.8×10^4

Sixth grade Chinese composition,
Reading "one night's work" has the feeling, good adds 100, bad adds 50

Today, I read the article "one night's work" with great excitement. This article tells us how hard Premier Zhou worked for the country and the people all night, but his daily life is so simple. Premier Zhou's work is very hard, and he has to correct "one foot high" one night

As shown in the figure, it is known that the function expression of the straight line L is y = - 43x + 8, and l intersects with the X axis and Y axis at two points a and B respectively. The moving point Q starts from point B and moves to point a at the speed of 2 unit lengths per second on the line Ba, while the moving point P starts from point a and moves to point o at the speed of 1 unit length per second on the line Ao. Let the time for the moving points P and Q be T seconds. (1) when t is the value, the △ Apq is Isosceles triangle based on PQ? (2) Find out the coordinates of points P and Q; (expressed by the formula containing T) (3) when the value of T is, the area of △ Apq is 15 times that of △ abo?

(1) When AQ = AP, it is an isosceles triangle with PQ as the base, the function expression of ∵ line L is y = − 43x + 8, and l and x-axis, Y-axis intersect at two points a and B respectively, ∵ a (6,0), B (0,8), ∵ AB = 10, ∵ AQ = 10-2t, AP = t, i.e., 10-2t = t, ∵ t = 103 (seconds), when t = 103, it is an isosceles triangle with PQ as the base; (2) through q-point, it leads a vertical line to x-axis and y-axis respectively, and its foot is perpendicular Let Q (x, y) BQ = 2T, AP = t (Q (x, y) BQ = 2T, AP = t (Q (x, y) BQ = 2T, AP = t, and △ NQN \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\area = 12 × 6 × 8 = 24  1 The solution of 2T × 45 (10 − 2t) = 15 × 24 is that T1 = 2, T2 = 3. When T1 = 2 seconds or T2 = 3 seconds, the area of △ Apq is 15 times that of △ abo

On the blackboard are written the seven numbers of 8, 9, 10, 11, 12, 13 and 14. Erase two numbers at any time, and then write the two numbers
How many times will there be only one number left on the blackboard? What is the number?

6 times (8 + 14) × 7 / 2 + 6 = 83

Why in the experiment of measuring the oxygen content in the air, the volume of water is the volume of oxygen in the air
What's the relationship between water and oxygen

In this experiment, red phosphorus reacts with oxygen to consume oxygen, while other substances will not react. Therefore, due to the pressure difference between the inside and outside of the device, water is pressed into the device to replace the volume of oxygen, so the volume of water inflow is the volume of oxygen in the air

As shown in the figure, in square ABCD, e is the point on AB, f is the point on BC, and AE + CF = EF

It is proved that when △ DAE is rotated counterclockwise around point d to the position of △ DCG, there is de = DG & nbsp; & nbsp; & nbsp; & nbsp; AE = CG & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ade = ∠ CDG, because AE + CF = EF  CG + CF = EF, then GF = EF, and DF = DF △ DGF & # 8773; △ def (SSS) ≠ ∠ GDF = ∠ EDF, then ∠ CDG +

3.6.4.10 and 3.5.11.7
3.6.4.10 write three
3.5.11.7 write one

First:
1:3 × 6 - 4 + 10
2:(3 × 6) - 4 + 10
3:(3 × 6 - 4) + 10
4:((3 × 6) - 4) + 10
5:3 × 6 -(4 - 10)
6:(3 × 6) - (4 - 10)
7:3 × 6 + 10 - 4
8:(3 × 6) + 10 - 4
9:(3 × 6 + 10) - 4
10:((3 × 6) + 10) - 4
11:3 × 6 +(10 - 4)
12:(3 × 6) + (10 - 4)
13:3 × (4 - 6 + 10)
14:3 × ((4 - 6) + 10)
15:3 × (4 - (6 - 10))
16:3 × (4 + 10 - 6)
17:3 × ((4 + 10) - 6)
18:3 × (4 + (10 - 6))
19:3 × (10 - 6 + 4)
20:3 × ((10 - 6) + 4)
21:3 × (10 - (6 - 4))
22:3 × (10 + 4 - 6)
23:3 × ((10 + 4) - 6)
24:3 × (10 + (4 - 6))
25:(3 × (10 - 4)) + 6
26:3 × (10 - 4) + 6
27:6 - (3 × (4 - 10))
28:6 - 3 ×(4 - 10)
29:6 × 3 - 4 + 10
30:(6 × 3) - 4 + 10
31:(6 × 3 - 4) + 10
32:((6 × 3) - 4) + 10
33:6 × 3 -(4 - 10)
34:(6 × 3) - (4 - 10)
35:6 + (3 × (10 - 4))
36:6 + 3 ×(10 - 4)
37:6 × 3 + 10 - 4
38:(6 × 3) + 10 - 4
39:(6 × 3 + 10) - 4
40:((6 × 3) + 10) - 4
41:6 × 3 +(10 - 4)
42:(6 × 3) + (10 - 4)
43:6 ÷ 3 × 10 + 4
44:(6 ÷ 3) × 10 + 4
45:(6 ÷ 3 × 10) + 4
46:((6 ÷ 3) × 10) + 4
47:(6 ÷ (3 ÷ 10)) + 4
48:6 ÷ (3 ÷ 10) + 4
49:6 - (4 - 10) × 3
50:6 - ((4 - 10) × 3)
51:6 × 10 ÷ 3 + 4
52:(6 × 10) ÷ 3 + 4
53:(6 × 10 ÷ 3) + 4
54:((6 × 10) ÷ 3) + 4
55:(6 × (10 ÷ 3)) + 4
56:6 × (10 ÷ 3) + 4
57:6 + (10 - 4) × 3
58:6 + ((10 - 4) × 3)
59:4 + 6 ÷ 3 × 10
60:4 + (6 ÷ 3) × 10
61:4 + (6 ÷ 3 × 10)
62:4 + ((6 ÷ 3) × 10)
63:4 + (6 ÷ (3 ÷ 10))
64:4 + 6 ÷(3 ÷ 10)
65:4 + 6 × 10 ÷ 3
66:4 + (6 × 10) ÷ 3
67:4 + (6 × 10 ÷ 3)
68:4 + ((6 × 10) ÷ 3)
69:4 + (6 × (10 ÷ 3))
70:4 + 6 ×(10 ÷ 3)
71:(4 - 6 + 10) × 3
72:((4 - 6) + 10) × 3
73:(4 - (6 - 10)) × 3
74:4 + 10 ÷ 3 × 6
75:4 + (10 ÷ 3) × 6
76:4 + (10 ÷ 3 × 6)
77:4 + ((10 ÷ 3) × 6)
78:4 + (10 ÷ (3 ÷ 6))
79:4 + 10 ÷(3 ÷ 6)
80:(4 + 10 - 6) × 3
81:((4 + 10) - 6) × 3
82:(4 + (10 - 6)) × 3
83:4 + 10 × 6 ÷ 3
84:4 + (10 × 6) ÷ 3
85:4 + (10 × 6 ÷ 3)
86:4 + ((10 × 6) ÷ 3)
87:4 + (10 × (6 ÷ 3))
88:4 + 10 ×(6 ÷ 3)
89:10 + 3 × 6 - 4
90:(10 + 3 × 6) - 4
91:(10 + (3 × 6)) - 4
92:10 + (3 × 6) - 4
93:10 + (3 × 6 - 4)
94:10 + ((3 × 6) - 4)
95:10 ÷ 3 × 6 + 4
96:(10 ÷ 3) × 6 + 4
97:(10 ÷ 3 × 6) + 4
98:((10 ÷ 3) × 6) + 4
99:(10 ÷ (3 ÷ 6)) + 4
100:10 ÷ (3 ÷ 6) + 4
101:10 + 6 × 3 - 4
102:(10 + 6 × 3) - 4
103:(10 + (6 × 3)) - 4
104:10 + (6 × 3) - 4
105:10 + (6 × 3 - 4)
106:10 + ((6 × 3) - 4)
107:10 × 6 ÷ 3 + 4
108:(10 × 6) ÷ 3 + 4
109:(10 × 6 ÷ 3) + 4
110:((10 × 6) ÷ 3) + 4
111:(10 × (6 ÷ 3)) + 4
112:10 × (6 ÷ 3) + 4
113:(10 - 6 + 4) × 3
114:((10 - 6) + 4) × 3
115:(10 - (6 - 4)) × 3
116:10 - 4 + 3 × 6
117:(10 - 4) + 3 × 6
118:10 - 4 +(3 × 6)
119:(10 - 4) + (3 × 6)
120:10 - (4 - 3 × 6)
121:10 - (4 - (3 × 6))
122:(10 - 4) × 3 + 6
123:((10 - 4) × 3) + 6
124:(10 + 4 - 6) × 3
125:((10 + 4) - 6) × 3
126:(10 + (4 - 6)) × 3
127:10 - 4 + 6 × 3
128:(10 - 4) + 6 × 3
129:10 - 4 +(6 × 3)
130:(10 - 4) + (6 × 3)
131:10 - (4 - 6 × 3)
132:10 - (4 - (6 × 3))
the second:
1:(3 - 7) × (5 - 11)
2:(5 - 11) × (3 - 7)
3:(11 - 5) × (7 - 3)
4:(11 × 7 - 5) ÷ 3
5:((11 × 7) - 5) ÷ 3
6:(7 - 3) × (11 - 5)
7:(7 × 11 - 5) ÷ 3
8:((7 × 11) - 5) ÷ 3

Each letter represents a number, then a = () B = () C = () d = () a B C multiplied by C equals DBC

125*5=625

In the quadrilateral ABCD, CE bisects ∠ BCD, f is the midpoint of AB, ab = 6, BC = 4, and the value of AE: EF: BF is calculated
2. It is known that the circumference of the quadrilateral ABCD is 36cm, and the heights of the two adjacent sides are 4cm and 5cm respectively
3. In quadrilateral ABCD, AC and BD intersect at O, ad = 8, ab = 10, BD = 6, find the length of BC, CD, ob, OA and quadrilateral ABCD area

I've done the first one, and I don't know if there are no other ones. I'll give you what I wrote
1. ∵ CE bisection ∠ BCD
∴∠BCE=∠DCE
And the quadrilateral ABCD is a parallelogram
Therefore, DCE = bec
That is, BCE = bec
∴BC=BE=4
F is the midpoint of ab
∴BF=1/2AB=3
∴EF=BE-BF=1
∴AE=AB-BE=2
That is AE: EF: BF = 2:1:3

If M and N are positive integers and have (MX + a) (x-3a) = 3x & # 178; + nax-3a & # 178;, find the value of M and n

(mx+a)(x-3a)
=mx^2-3max+ax-3a^2
=mx^2+a(1-3m)x-3a^2
=3x^2+nax-3a^2
m=3
n=1-3m=1-3*3=-8