70 two part operations in multiplication and division of rational numbers

70 two part operations in multiplication and division of rational numbers

（1）(-9)-(-13)+(-20)+(-2)＝
(2) 3+13-(-7)÷6＝
(3) (-2)-8-14-13＝
(4) (-7)×(-1) ÷7+8 ＝
(5) (-11)×4-(-18) ÷18 ＝
(6) 4+(-11)-1÷(-3) ＝
(7) (-17)-6-16÷(-18) ＝
(8) 5÷7+(-1)-(-8) ＝
(9) (-1)×(-1)+15+1 ＝
(10) 3-(-5)×3÷(-15) ＝
(11) 6×(-14)-(-14)+(-13) ＝
(12) (-15)×(-13)-(-17)-(-4) ＝
(13) (-20) ÷13÷(-7)+11 ＝
(14) 8+(-1) ÷7+(-4) ＝
(15) (-13)-(-9)×16×(-12) ＝
(16) (-1)+4×19+(-2) ＝
(17) (-17)×(-9)-20+(-6) ＝
(18) (-5) ÷12-(-16)×(-15) ＝
(19) (-3)-13×(-5)×13＝
(20) 5+(-7)+17-10 ＝
(21) (-10)-(-16)-13×(-16) ＝
(22) (-14)+4-19-12 ＝
(23) 5×13÷14÷(-10) ＝
(24) 3×1×17÷(-10) ＝
(25) 6+(-12)+15-(-15) ＝
(26) 15÷9÷13+(-7) ＝
(27) 2÷(-10)×1-(-8) ＝
(28) 11÷(-19)+(-14)-5 ＝
(29) 19-16+18÷(-11) ＝
(30) (-1)÷19+(-5)+1 ＝
(31) (-5)+19÷10×(-5) ＝
(32) 11÷(-17)×(-13)×12 ＝
(33) (-8)+(-10)÷8×17 ＝
(34) 7-(-12)÷(-1)+(-12) ＝
(35) 12+12-19+20 ＝
(36) (-13)×(-11)×20+(-4) ＝
(37) 17÷(-2)-2×(-19) ＝
(38) 1-12×(-16)+(-9) ＝
(39) 13×(-14)-15÷20 ＝
(40) (-15)×(-13)-6÷(-9) ＝
(41) 15×(-1)÷12+7 ＝
(42) (-13)+(-16)+(-14)-(-6) ＝
(43) 14×12×(-20)×(-13) ＝
(44) 17-9-20+(-10) ＝
(45) 12÷(-14)+(-14)+(-2) ＝
(46) (-15)-12÷(-17)-(-3) ＝
(47) 6-3÷9÷(-8) ＝
(48) (-20)×(-15)×10×(-4) ＝
(49) 7÷(-2)×(-3)÷(-14) ＝
(50) 13÷2×18×(-7) ＝
(51) 13×5+6+3＝
(52) (-15)÷5÷3+(-20) ＝
(53) 19×4+17-4 ＝
(54) (-11)-(-6)×(-4)×(-9) ＝
(55) (-16)+16-(-8)×(-13) ＝
(56) 16÷(-1)÷(-10)÷(-20) ＝
(57) (-1)-(-9)-9÷(-19) ＝
(58) 13×20×(-13)×4 ＝
(59) 11×(-6)-3+18 ＝
(60) (-20)+(-12)+(-1)+(-12) ＝
(61) (-19)-3×(-13)×4 ＝
(62) (-13)÷3-5×8 ＝
(63) (-15)÷1+17×(-18) ＝
(64) (-13)÷3÷19÷8 ＝
(65) (-3)÷(-13)÷20×5 ＝
(66) 3÷12÷(-18)-18 ＝
(67) 5×(-19)÷13+(-6) ＝
(68) 4+4×(-19)-11 ＝
(69) (-2)+17-5+(-1) ＝
(70) 9+(-3)×19×(-19) ＝
(71) (-12)-(-6)+17÷2 ＝
(72) 15×(-5)-(-3)÷5 ＝
(73) (-10)×2÷(-1)÷4 ＝
(74) (-8)×16÷(-6)+4＝
(75) 2-11+12+10 ＝
(76) (-3)+(-20)×(-7)×(-9) ＝
(77) (-15)+8-17÷7 ＝
(78) (-14)×10+18×2 ＝
(79) (-7)+2-(-17)×19 ＝
(80) (-7)÷18÷1+1 ＝
(81) 11÷(-9)-(-16)÷17 ＝
(82) 15+5×6-(-8) ＝
(83) (-13)×(-18)+18÷(-6) ＝
(84) 11-(-1)÷11×(-6) ＝
(85) (-4)+(-12)+19÷6 ＝
(86) (-18)÷(-1)÷(-19)+2 ＝
(87) 9×(-8)×(-6)÷11 ＝
(88) 20×(-3)×(-5)+1 ＝
(89) (-18)-2+(-11)÷20 ＝
(90) 15×1+4×17 ＝
(91) 1-10+(-14)÷(-1) ＝
(92) 10+(-4)×(-19)+(-12) ＝
(93) 15÷14÷5×7 ＝
(94) 8+(-13)÷3+1 ＝
(95) (-14)+6+(-2)×(-14) ＝
(96) (-5)÷(-13)÷4+7 ＝
(97) (-15)÷(-2)÷(-12)+(-2) ＝
(98) (-17)-(-20)-20×(-10) ＝
(99) (-7)-10-13÷3 ＝
(100) (-20)+(-18)+11+9＝
1 -18
2 103÷6
3 -37
4 9
5 -43
6 -(20÷3)
7 -(199÷9)
8 54÷7
9 17
10 2
11 -83
12 216
13 1021÷91
14 27÷7
15 -1741
16 73
17 127
18 -(2885÷12)
19 842
20 5
21 214
22 -41
23 -(13÷28)
24 -(51÷10)
25 24
26 -(268÷39)
27 39÷5
28 -(372÷19)
29 15÷11
30 -(77÷19)
31 -(29÷2)
32 1716÷17
33 -(117÷4)
34 -17
35 25
36 2856
37 59÷2
38 184
39 -(731÷4)
40 587÷3
41 23÷4
42 -37
43 43680
44 -22
45 -(118÷7)
46 -(192÷17)
47 145÷24
48 -12000
49 -(3÷4)
50 -819
51 74
52 -21
53 89
54 205
55 -104
56 -(2÷25)
57 161÷19
58 -13520
59 -51
60 -45
61 137
62 -(133÷3)
63 -321
64 -(13÷456)
65 3÷52
66 -(1297÷72)
67 -(173÷13)
68 -83
69 9
70 1092
71 5÷2
72 -(372÷5)
73 5
74 76÷3
75 13
76 -1263
77 -(66÷7)
78 -104
79 318
80 11÷18
81 -(43÷153)
82 53
83 231
84 115÷11
85 -(77÷6)
86 20÷19
87 432÷11
88 301
89 -(411÷20)
90 83
91 5
92 74
93 3÷2
94 14÷3
95 20
96 369÷52
97 -(21÷8)
98 203
99 -(64÷3)
100 -18

One by half plus two by three plus three by four Add eight times nine and nine times ten

2010 Shanghai World Expo makes us full of expectations. By that time, rail transit will account for 50% of passenger traffic, public bus will account for 30%, and water transport will account for 5%
The 2010 Shanghai World Expo makes us full of expectations. At that time, rail transit accounts for 50% of the passenger traffic, public bus accounts for 30% and water transport accounts for 5%. It is predicted that 72.2 million people will attend the Expo. How many visitors will be transported by rail transit, public bus and water transport?

Then make sure your simple answer is yes

Density of steel

The density of steel is 7.85g/cm3
Calculation of theoretical weight of steel
The unit of measurement for calculating the theoretical weight of steel is kilogram (kg)
W (weight, kg) = f (sectional area, mm2) × L (length, m) × ρ (density, g / cm3) × 1 / 1000
The formula for calculating the theoretical weight of various steels is as follows:
Name (unit)
Calculation formula
Symbolic meaning
Calculation examples
Round steel wire rod (kg / M)
　　W= 0.006165 ×d×d
D = diameter mm
For round steel with diameter of 100 mm, calculate the weight per M. weight per M = 0.006165 × 1002 = 61.65kg
Rebar (kg / M)
　　W= 0.00617 ×d×d
D = section diameter mm
The weight per m of rebar with cross section diameter of 12 mm = 0.00617 × 12 2 = 0.89kg
Square steel (kg / M)
　　W= 0.00785 ×a ×a
A = edge width mm
For square steel with side width of 20 mm, calculate the weight per M. the weight per M = 0.00785 × 202 = 3.14 kg
Flat steel
　　（kg/m）
　　W= 0.00785 ×b ×d
B = side width mm
D = thickness mm
For flat steel with side width of 40 mm and thickness of 5 mm, calculate the weight per M. the weight per M = 0.00785 × 40 × 5 = 1.57 kg
Hexagons
　　（kg/m）
　　W= 0.006798 ×s×s
S = opposite distance mm
For the angle steel with 50 mm distance from the opposite side, calculate the weight per M. the weight per M = 0.006798 × 502 = 17kg
Octagonal steel
　　（kg/m）
　　W= 0.0065 ×s ×s
S = opposite distance mm
For octagonal steel 80 mm apart from the opposite side, calculate the weight per M. the weight per M = 0.0065 × 802 = 41.62kg
Equilateral angle steel
　　（kg/m）
　　= 0.00785 ×[d （2b – d ）+0.215 （R2 – 2r 2 ）]
B = side width
D = edge thickness
R = inner arc radius
R = end arc radius
Find out the weight per m of 20 mm × 4 mm equilateral angle steel. Find out from the metallurgical product catalog that R of 4 mm × 20 mm equilateral angle steel is 3.5, R is 1.2, then the weight per M = 0.00785 × [4 × (2 × 20 – 4) + 0.215 × (3.52 – 2 × 1.22)] = 1.15 kg
Unequal angle steel
　　（kg/m）
　　W= 0.00785 ×[d （B+b – d ）+0.215 （R2 – 2 r 2 ）]
B = length side width
B = short side width
D = edge thickness
R = inner arc radius
R = end arc radius
The weight per m of 30 mm × 20 mm × 4 mm unequal angle steel is calculated. It is found from the metallurgical product catalog that R of 30 × 20 × 4 unequal angle steel is 3.5 and R is 1.2, then the weight per M = 0.00785 × [4 × (30 + 20 – 4) + 0.215 × (3.52 – 2 × 1.22)] = 1.46kg
Channel steel
　　（kg/m）
　　W=0.00785 ×[hd+2t （b – d ）+0.349 （R2 – r 2 ）]
H = high
B = leg length
D = waist thickness
T = average leg thickness
R = inner arc radius
R = end arc radius
Find out the weight per m of 80 mm × 43 mm × 5 mm channel steel. Find out from the metallurgical product catalog that the channel steel t is 8, R is 8, R is 4, then the weight per M = 0.00785 × [80 × 5 + 2 × 8 × (43 – 5) + 0.349 × (82 – 42)] = 8.04kg
I-beam (kg / M)
　　W= 0.00785 ×[hd+2t （b – d ）+0.615 （R2 – r 2 ）]
H = high
B = leg length
D = waist thickness
T = average leg thickness
R = inner arc radius
R = end arc radius
The weight per m of 250 mm × 118 mm × 10 mm I-beam is calculated. According to the manual of metal materials, if t is 13, R is 10 and R is 5, the weight per M = 0.00785 × [250 × 10 + 2 × 13 × (118 – 10) + 0.615 × (102 – 52)] = 42.03 kg
Steel plate (kg / m2)
　　W= 7.85 ×d
D = thickness
For 4mm thick steel plate, the weight per m2 is calculated. The weight per m2 is 7.85 × 4 = 31.4kg
Steel pipe (including stainless steel pipe)
Seam steel pipe and welding
Steel pipe (kg / M)
　　W= 0.02466 ×S （D – S ）
D = outer diameter
S = wall thickness
For seamless steel pipe with outer diameter of 60 mm and wall thickness of 4 mm, calculate the weight per M. the weight per M = 0.02466 × 4 × (60 – 4) = 5.52kg

There was a batch of cement on the construction site. After two fifths of the cement was used, another 30 tons were transported. At this time, the cement was 20% more than the original

What is your question? If the question is how much cement there is, you can ask this:
Let the unknown be x, X-2 / 5x + 30 = x + 1 / 5x, x = 50

The volume and height of a cylinder and a cone are equal. The circumference of the bottom surface of the cone is 12.56 decimeters. The height is 3 decimeters. What is the bottom area of the cylinder?

Bottom area of cylinder X3 = volume of cone = (1 / 3) (12.56 / 2 / 3.14) x (12.56 / 2 / 3.14) x3.14
Bottom area of cylinder = 1.39555 square decimeter

A number is not only a multiple of 9, but also a factor of 54. What is the possible number

9.18.27.54

When x and Y take what value, the value of x ^ 2-3xy + 2Y ^ 2-4x + 5y-9 is the smallest
^The sign is the square of X! I think the minimum value is to match the complete square, but this problem seems to be difficult to match. If we decompose the factor, I don't know how to solve it?

2x-3y-4=0
-3x+4y+5=0
x=-1
y=-2
here
Original formula = 1-6 + 8 + 4-10-9 = - 12

6×（_ -3) = 2 × () if () is the same, how to calculate it? Can you say the basic principle and its formula in the most popular language
Can it be calculated as follows: 6 × (_ -3）=2×（）
Namely: 6 × () - 18 = 2 × ()
6×（）-2×（）
4×（）=18
18 △ 4 = 4.5, please correct? Would you please say the basic principle and its formula in the most popular language

6×（4.5-3）=2×（4.5）
Let this number be X
6(X-3)=2X
6X-18=2X
4X=18
X=4.5

A polynomial minus X & # 178; - Y & # 178; equals x & # 178; + Y & # 178;, then the polynomial is

This polynomial is (X & # 178; + Y & # 178;) + (X & # 178; - Y & # 178;)
=2x²