1. Multiple choice problem
1A 2D 3A 4C
fill (up) a vacancy
( 1)T=20-6h 20，6 T h h
(2)Q = 6x 105-pt 6x 105 p Q t 0≤t≤6x 105/p
(3)S= 1.5b (4) 0≤x≤7 0≤y≤5 5 0
solve problems
( 1)y = Q/a-x–Q/a(0≤x≤a)
(2)y=80-2x
20 & ltx & lt40
(3) ①-2≤x≤3
② When x=3, the minimum value of y is 1/2.
③ When -2≤x≤0, Y increases with the increase of X, and when 0≤x≤3, Y decreases with the increase of X..
(4)①`v=800-50t
②0≤t≤ 16
③ When t=8, v=800-50x8=400.
④ When v= 100, 100=800-50t.
T= 14
Page 5 to Page 7
Multiple choice
1B 2C 3C 4B 6A 7B 8D
fill up a vacancy
( 1) 1(2)y = 2x+ 1- 1(3)m & lt； 2n & lt； 3 (4)y=-3x+3
(5)y = x+3(6)y = 64x+48(7)S = 2n+ 1(8)y = 1/5 x-6 30
answer the question
(1) let y = kx+b.
-4k+b= 15
6k+b= -5
k= -2 b=7
y= -2x+7
(2) Omission
(3)① indicates the relationship between y and x, and x is the independent variable.
② Leaving home1010km1330km.
③ 30km away from home12-13.
④13km
⑤2: 00 - 13: 00
615km/h
Page 9-Page 1 1
1. Multiple choice problem
( 1)A (2)C (3)C
fill (up) a vacancy
( 1)y =-2x(2)m & lt； 2(3)y = 5x+3(4)y2 & gt； y 1(5)y =-2x+ 100 25 & lt； x & lt50
(6)9
solve problems
( 1) ① Q=200+20t② (0≤t≤30)
(2) ①y=80 (0≤x≤50)
y = 1.9x- 15(50≤x≤ 100)
②y= 1.6x
③ Option 1
(3)① y=25/72x on the same straight line
② When x=72, y=25.
When x= 144, y=50.
When x=2 16, y=75.
y=25/72 x (0≤x≤345.6)
③ When x= 158.4, y=25/72x 158.4=55.
(4) ①y A =2x+ 180
Y B = 2.5x+ 140
② when x= 100, Y A =200+ 180=380.
Y B = 140+250=390
380〈390
Renting armored vehicles is more flexible.
Page 13-Page 15
1. Multiple choice problem
( 1)D (2)C (3)C
fill (up) a vacancy
( 1)x=2
y=3
(2)x = 2x & gt； 2
(3)-3 -2 x= -5/8 y= - 1/8
(4) 1/2 0 x=2
y=3
(5)y=5/4 x
solve problems
3.( 1) Omit
(2)① According to the meaning of the question
-k+b= -5
2k+b= 1
solve
k=2 b= -3
y=2x+3
When y≥0
2x-3≥0，x≥3/2
② when x
2x-3
That is y < 1
(3) ①y membership card =0.35+ 15
Y book card = 0.5x
② If Y membership card < Y book rental card
Then 0.35x+ 15
x & gt 100
If the book is rented for more than 100 days, the membership card is more cost-effective than the book rental card.
(4) let A(m, n)
1/2x4xm=6
m=3
n=2
One (-3, -2)
y=2/3x，y= -2/3x -4
(5)①y A = 0.8x 1.5x+900 = 1.2x+900(x≥500)
y B = 1.5x+900 x 0.6 = 1.5x+540(x≥500)。
② if Y A =y B
1.2x+900= 1.5x+540
x= 1200
When x
When x= 1200, the charges of the two factories are the same.
When x > 1200, select a factory.
2000 > 1200, factory selection.
Y A = 1.2x2000+900=3300
This answer is provided free of charge by Wuhan Junior High School Science Network (www.wh11.com), and it is strictly prohibited to reprint it on other websites.
Page 17-Page 19
1. Multiple choice problem
( 1)C (2)D (3)C
fill (up) a vacancy
( 1) 630 (2) 0. 17 0. 17 (3) 35 (4) ① 238. 1824 ② 129 ③ 20,000.
3 solve the problem
( 1)
Asia is located on seven continents.
② Asia and Africa
③ 100%
④ Oceania
⑤ No.
(2) (1) Workshop in the fourth quarter
② Workshop 1 and Workshop 2
③ ① from figure (1) and ② from figure (2).
(3) 148 20.25 ③ Which score has the most students? 70.5~80.5 students are the most.
Page 2 1-Page 23
1. Multiple choice problem
( 1)B (2)B (3)C (4)B
fill (up) a vacancy
(1)20% 30% 25% 25% (2) 36% 1 15.2 degrees (3)4 1 1.
3 solve the problem
( 1)
county
Population (ten thousand) 9015 72 273 771
Percentage12.9% 2.1%10.3% 39.1%1.0% 24.5%
Central angle 46.4 7.7 37.1140.8 39.7 88.2
(2) Sketch
(3) Height frequency (cm)
154.5~ 159.5 2
159.5~ 164.5 4
164.5~ 169.5 6
169.5~ 174.5 10
174.5~ 179.5 5
179.5~ 184.5 3
(4) Sketch conclusion: Only a few people are dissatisfied with their work.
answer
gruel of millet flour and sugar
3 fans
second floor
(5)① 20 0. 16 ② omitted
Pages 25 to 27
1. Multiple choice problem
( 1)B (2)C (3)A(4)C (5)B (6)C
fill (up) a vacancy
( 1)∠D∠C DC OD OC(2)D E C DE∠D 600
(3)CAD CD(4)500 10 10 8(5)ADE CAE
3 solve the problem
(1) ①△DCE can be regarded as △ABF translation and rotation.
②AF is not necessarily parallel to DE, because ∠AFE is not necessarily equal to ∠ D.
(2)∠ABC = 1800 X5/ 18 = 500
∠C= 1800x3/ 18=300
∠B'CB=∠A+∠ABC=800
∫△ABC?△A ' b ' c '
∴∠A'=∠A=300
∠B'=∠ABC=500
∠B'BC= 1800-∠B'-∠B'CB=500
(3)① Slightly ② Take the midpoint of each side and connect them in pairs.
(4) extend AD to e, make AD=DE, and connect BE.
∴ AD=ED
D is the midpoint of BC
In △BDE and △CDA,
BD=CD ∠ADC=∠BDE DE=DA
∴△BDE≌△CDA
∴BE=AC
AE & lta b+ BE 2AD & lt； AB+AC
∴ad<； AB+AC/2
Page 29-Page 365438 +0
Multiple choice
( 1)D (2)B (3) B (4)C
fill (up) a vacancy
( 1)6(2)200(3)BO = CO(4)AB = DC∠ACB =∠DBC
solve problems
( 1) ∵AE=CF
∴AE+EF=CF+EF
∴AF=CE
CD = AB DE = BF CE = AF
∴△CDE≌△ABF
∴∠DEC=∠AFB
∴DE‖BF
(2)△ Abe △ ACG
△ABD?△ACF
AB = AC
∴∠ABC=∠ACB
BD shares ∠ABC, CF shares ∠ACB.
∴∠ABD=∠ACF