I think you wrote it wrong? It should be the second volume of ninth grade mathematics, right? These questions can't be found in the first volume of ninth grade mathematics?

The following are the answers to the questions you mentioned in the second volume of ninth grade mathematics of People's Education Press:

P3 1 1—7

1, according to the meaning of the question, AE=4-x, eg = 4+X.

∴y=(4-x)(4+x)=-x？ + 16(0 & lt； x & lt4)

2. According to the question, the sales volume in the second year is 5000( 1+x), and the sales volume in the third year is 5000( 1+x)? Taiwan Province, y=5000(x+ 1)?

3、D

Step 4 sketch

( 1)y=x？ +2x-3, opening direction upward, symmetry axis x=- 1, vertex coordinate (-1, -4).

(2)y= 1+6x-x？ , the opening direction is downward, the symmetry axis x=3, and the vertex coordinate (3, 10).

(3)y= 1/2x？ +2x+ 1, the opening direction is upward, the symmetry axis x=-2, and the vertex coordinates (-2,-1).

(4)y=- 1/4x？ +x-4, opening direction downward, symmetry axis x=2, vertex coordinate (2, -3)

5、∫s = 15t-6t？ =-6(t-5/4)？ +75/8

When t=5/4, the maximum value of S is 75/8.

After the car braked, it stopped and moved 75/8m.

6 、( 1)y=7/8x？ +2x+ 1/8

(2)y=20/3x？ -20/3x-5

7. If the length is x meters, the width is (30-x)/2m.

The area of vegetable garden can be expressed as y=x( 15-x/2)=-(x? /2)+ 15x =- 1/2(x- 15)？ + 1 12.5

When x= 15, the maximum value of y is 1 12.5.

∴ When the rectangle is 15m long and 7.5m wide, the vegetable garden area is the largest, and the largest area is 1 12.5m?

p32 8-9

8. When s=85, 85= 1.8t+0.064t? T=25, so it took him 25 seconds to cross this hillside.

9. If the length of a rectangle is x cm and the width is (36-2x)/2=( 18-x) cm.

Lateral area y=2πx ×( 18-x)=-2π(x-9) when a cylinder rotates around one side? + 162π

When x=9, the lateral area is the largest, that is, when the length and width of the rectangle are 9cm, the lateral area of the cylinder is the largest.

P70 1—6

1, ∫ Similar polygons have equal corresponding angles and equal proportions of corresponding edges.

∴∠E=∠K,∠G=∠M,∠F=∠360 -(∠E+∠H+∠G)，∠F=∠N

∴∠E=67，∠G= 107，∠n = 360-(67+ 107+ 143)= 43

∵x/35=6/y= 10/z=4/ 10,∴x= 14,y= 15,x=25

2.∵ The ratio of the corresponding sides of similar triangles is equal. Let the other two sides of △DEF be x and y, respectively, and the circumference be c.

∴5/ 15= 12/x= 13/y,c= 15+x+y

∴x=36,y=39,C=90

3 、( 1)≈∠ 1 =∠2，∠g =∠I = 90 ,∴△fgh∽△jih,∴3/6=x/8=5/y,∴x=4,y= 10

(2)≈fhg+∠ghj =∠khj+∠khf，∠KHF=∠GHJ=90 ,∴∠GHF=∠KHJ

* GH/KH = FH/HJ = 3/2，∴△GFH∽△KHJ，∴ X = 124，y/22=3/2，∴y=33.

4. The area ratio is equal to the square of the side length ratio.

∴ If the advertising area becomes 9 times of the original, the advertising fee is 180×9= 1620 (yuan).

Step 5 sketch

Firstly, the center o of similarity is selected, and then it is drawn according to the characteristics of similarity map.

6. According to the similarity nature, the similarity ratio between the characters on the blackboard and the characters in the textbook is 6: 0.3 = 20: 1.

If the words on the blackboard are x cm long and y cm wide, students can feel the same as the words in the textbook when reading.

x/0.4=y/0.35=20/ 1，x=8，y=7

∴: The words on the blackboard should be 7 cm× 8 cm.

P7 1 7— 10

7、∵oa/oc=ob/od,∠doc=∠aob,∴△doc∽△aob

∴CD/AB=OC/OA, that is, B/AB =1N, ∴AB=nb, ∴x= 1/2(a-nb).

8.∫c is a point on the circumference, ∴ ∠ ACB = 90.

∴∠ACP+∠PCB=90

∴∠∵cd⊥ab Polychlorinated Biphenyls+∠ Polybrominated Biphenyls = 90.

∴∠ACP=∠BPC

∠∠APC =∠BPC = 90。

∴△APC∽△CPB,∴PA/PC=PC/PB,∴PC？ =PA×PB

9, the process is slightly, the ball can touch the wall 5.4m high from the ground.

10, 35mm=0.035m, 50mm=0.05m, 70mm=0.07m According to the meaning of the question, △XYL∽△ABL.

When the focal length is 50mm, 0.035 m/ab = 0.07 m/5 m.

∴AB=2.5m

So when the focal length is 70mm, the scene that can be photographed at 5m is 2.5m wide.

P72 1 1— 12

1 1、∵db‖ac,∴△dob∽△coa,∴od/oc=ob/oa,∴oa×od=ob×oc

12, let the width of the shadow part be x cm and the length of the shadow part be 6cm.

The original rectangle is similar to the shaded part.

∴ 10/6=6/x,∴x=3.6

∴ The rectangular area on the left is S=3.6×6=2 1.6cm?

P97 1—9

1, ∫ in Rt△ABC, ∠ C = 90, a=2, Sina = 1/3, ∴c=a/sinA=2/( 1/3)=6.

∴b=√6？ -2? =4√2

∴cosa=b/c=(4√2)/6=(2√2)/3,tana=a/b=2/(4√2)=(√2)/4

2、≈c = 90 ,cosa=(√3)/2,∴ac/ab=(√3)/2

∫AC = 4√3，∴AB=(4√3)/(√3/2)=8.

∴BC=√8？ -(4√3)? =4

3.( 1) Original formula =√2×(√2)/2- 1=0

(2) The original formula =√3×(√3/2)+√3-2×(√3/2)? =3/2+√3-2×(3/4)=√3

4、( 1)0.54 (2)0.43 (3)7.27 (4)-0.04

5 、( 1)A = 40.08(2)A = 69. 12(3)A = 88.38(4)A = 35.26

6、

(1) If the vertex angle is 30 and the waist is 2√3, AB=AC=2√3, BC = 2 × AC × COS75 = 4 √ 3 COS75.

The circumference of ABC is AB+AC+BC ∴ 8.6.

(2) If the vertex angle is 30 and the bottom edge is 2√3, BC=2√3, AB = AC = (√ 3)/COS 75.

The circumference of ABC is AB+AC+BC≈ 16.8.

(3) If the vertex angle is 30, the waist is 2√3, AB=AC=2√3, BC = 2 ABCOS 30 = 4 √ 3× (√ 3/2) = 6.

The circumference of ABC is AB+AC+BC=6+4√3.

(4) If the base angle is 30 and the bottom edge is 2√3, BC=2√3, and AB=(√3)/(√3/2)=2=AC.

The circumference of ABC is AB+AC+BC=4+2√3.

7. The process is short, and the distance between the ship and the coast is 42/tan 33 ≈ 65m.

8. Tan 43° 24' = ab/BC, ∴ AB = BC× Tan 43 24' ≈ 30.8m.

Let DE⊥AB be at point E when crossing point D, ∫tan 35 12' = AE/de, AE = DE× TAN3512' ≈ 23.0m.

∴DC=AB-AE=30.8-23.0=7.8m, so the heights of these two buildings are 30.8m and 7.8m respectively.

9. Make CG⊥CD and meet the BA extension line at G point; For BF⊥AB, the CD extension line is at f; Draw D as DE⊥AB and E as a cross.

∫ ≈ EDB = 30, ∴ ∠ DBF = 30, Ag = CG = BF = 5cm, ∴ BD = BF/COS30 =10/.732 ≈ 5.8m.

DF=5/√3≈2.9，∫∠GCA = 45 ,∴ac=5/(√2/2)=5√2≈7.3m

∴ab=cf-ag=3.4+5/√3-5= 1.3m

P98 10— 13

10, (1) 5.8m (2) 66, so this ladder can be used safely.

1 1 、( 1)△AFB∽△FEC

(2) Let CE=3x and CF=4x, then AB=8x, BF=6x, AF= 10x. In Rt△AEF, AF? +EF？ =AE？

∴(5x)？ +( 10x)？ =(5√5)? X= 1, and the circumference is 2( 10x+8x)=36cm.

12, knowing AB, BC and their included angle ∠B, we can find the area s of parallelogram ABCD.

S=AB×BC×sin∠B

13、

(1) The perimeter of the inscribed regular N-polygon is 2 nrsin (180/n).

What is the inscribed area of a regular N-polygon? Sine (180 /n) cosine (180 /n)

(2)

Inscribed regular polygon, regular hexagon, regular dodecagon and regular icosahedron. ...

Circumference 6r6.2 1r6.26r ...

Area 2.6R? 3R？ 3. 1R？ ……

P 125 1—3

1. The orthographic view corresponding to the three views in the figure is (3).

2, sketch (draw it yourself, it is not convenient to operate here)

3. There are three cubes on the bottom floor, two cubes on the second floor, which are misaligned with the cubes on the bottom floor 1/2, and a cube on the top floor is placed on the cube on the right side of the second floor.

p 126 4-7

Step 4 sketch

5. Regular hexagonal prism

6. Three Views

The object is a cylinder with a bottom radius of 5 and a height of 20.

∴ The volume is V=π×5? ×20=500π

The surface area is S=2π×5×20+2π×5? =250π

7. Expand the sketch

The surface area is S=π×(5√2)? ×( 1/√2)+20×2π×5+π×5? =25(√2 +9)π