Exercise Book-Math-Grade 9-Reference Answer

Chapter 1-Chapter 2

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Chapter 1 Inverse Proportional Function

1. 1( 1)

1. No, yes, yes, yes, no; /,3, 1/2,-π,/

2. All real numbers of 2.x ≠ 0, 1/4,-1.

3. The answer is not unique. If the resolution function is y = 12/x, there are: (1)3 (2)3/2 (3)-3/2.

(1) v = 240/t (2) when t = 3.2h, v = 75km/h.

5.( 1)S=600/x (2)a=300/b

6.( 1) A = 16/h, and h is a real number greater than 0.

(2) The sum of the upper and lower soles is 8cm, the waist circumference AB = CD = 2 √ 2 cm, and the trapezoid circumference is (8+4√2)cm.

1. 1(2)

1.- 12

2. all real numbers of y =10/x, x≠0.

3. y =-√ 6/X. When x = √ 6, y =- 1.

4.( 1)y=2z，z=-3/x

(2)x=-3/5，y= 10

(3) y =-6/x, yes.

5.( 1)D = 100/ sec

(2) 150 degrees

6.( 1) y = 48/x, yes, the practical significance of scale factor 48 is that the area of this group of rectangles is 48cm^2.

(2) Let one side of a rectangle be a(cm) and the other side be 3a(cm). Substituting x = a and y = 3a into y = 48/x to get a = 4, then the circumference of the rectangle is 2 (a+3a) = 32 (cm).

1.2( 1)

1.y=-√2/x

2.B

3.( 1) table omitted

(2) Sketch

4.( 1)y=4/x

(2) Sketch

5. The analytical formula of (1) inverse proportional function is y = 8/x, and the coordinates of one intersection point are (2,4) and the other intersection point is (-2,4).

6. According to the meaning of the question, {3m- 1 > 0, 1-m > 0, the solution is 1/3 < m < 1.

1.2(2)

1 .2, 4; expand

2.C

3.m 0。

(2) Omission

(3)2400 Pa, at least 0.1m 2.

7. two, four

8.a′(2，4)，m=8

9.( 1) From {-2k 2-k+5 = 4, k < 0 is k =- 1. Y = (- 1)/x。

(2)m= √3

10.( 1) Substitute P( 1, -3) into y =-(3m)/x, and get m = 1, then the analytical formula of the inverse proportional function is y =-3/x. Put the point p (/kloc-.

(2) Let y =-2x- 1 = 0, the abscissa of point p' is-1/2, and the area of △POP' is1/2 | |-3 | = 3/4.

1 1.( 1) If the coordinate of point A is (-1, a), the coordinate of point B is (1, -a). From the area of △ADB is 2, a = 2 can be obtained. Therefore, the analytical expressions of these two functions are

(2) Let AD be the base of △ADP, and when the abscissa of point P is -5 or 3, the area of △ADP is 4. , so the coordinates of point P are (3, -2/3), (-5, 2/5).

12. make the axis ab⊥x .∫ab = a "b" = | b |, Bo = b "o = | a |, ∴ RT △ ABO ≌ RT △ A" b "o, ∴ OA = OA

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Chapter II Quadratic Functions

2. 1

1.B

2.y=-x^2+25π

3. 1,-2,- 1; 3,0,5; - 1/2,3,0; 2,2,-4; 1,-2√2, 1

4.y=-2/3x^2+7/3x+ 1

5.( 1)s=- 1/2x^2+4x(0