Happy holidays, ninth grade math answers.
fill-in-the-blank question
1.(200 1 Changzhou, Jiangsu 1). Given x+y= 1, the value of the algebraic expression x3+3xy+y3 is ▲.
Answer 1.
Find the algebraic value of the test center.
Analysis as long as the required algebraic expression is converted into a known form, and then the known is converted into:
.
2. (Changzhou City, Jiangsu Province, 2002 1) If │ x │+3 = │ x-3 │, the value range of x is ▲.
The answer x≤0.
The nature of the absolute value of the test site.
Analysis According to the nature of absolute value, simplifying absolute value can analyze three situations: x≥3, 0≤x≤3 and x≤0:
① When x≥3, the original formula can be changed to: x+3 = x-3, with no solution;
② When 0≤x≤3, the original formula can be changed to: x+3 = 3-x, x = 0；;
③ When x≤0, the original formula can be simplified to -x+3 = 3-x, and the equation holds.
To sum up, the value range of x is x≤0.
(Changzhou City, Jiangsu Province, 2 minutes in 2003) The light shines on the plane mirror I as shown in the figure, and then reflects back and forth between the plane mirrors I and II. It is known that ∠ α = 60, ∠β = 50 and ∠ γ = ▲ degrees.
Answer 40.
Interdisciplinary problems in test sites, nature of reflection, definition of right angle, interior angle of triangle and theorem.
The angle between incident light and horizontal line is equal to the angle between reflected light and horizontal line, the definition of right angle and the theorem of triangle interior angle are analyzed and solved.
As shown in the answer sheet, according to the nature of reflection, we can get
∠BAC=∠α=60，∠ABC= 180 -2∠β=80，∠ACB=∠γ.
In △ABC, ∠ BAC+∠ ABC+∠ ACB = 180, then
∠ ACB = 180-(∠ BAC+∠ ABC) = 40, that is ∠ γ = 40.
4. (Changzhou City, Jiangsu Province, 2 points in 2004) As shown in the figure, point D is a point on the hypotenuse AB of Rt△ABC, with DE⊥BC in E and DF⊥AC in F. If AF= 15, BE= 10, the area of quadrilateral DECF is ▲.
Answer 150.
Determination and nature of rectangular test sites, parallelism, and determination and nature of similar triangles.
Analysis ∵DF⊥AC, DE⊥BC, ∴ DFC = ∠ C = ∠ DEC = 90, ∴ quadrilateral DFCE is a rectangle.
∴DF∥BC, and then ∠ ADF = ∠ B. And ∠AFD=∠DEB, ∴△ADF∽△DBE.
∴, that is, Germany? DF=AF？ BE= 150 .
∴ area of quadrilateral DFCE =DE? DF= 150 .
5. (Changzhou City, Jiangsu Province, 2005, 4 points) If part of the image of a parabola is shown in the figure, then the symmetry axis of the parabola is a straight line x= ▲, and the value range of X satisfying y < 0 is ▲. If a parabola is translated by ▲ units, a parabola is obtained.
Answer 3; 1< 1, where n is an integer) has a surface area of 6n2. If it is cut into n3 small cubes with a side length of 1, and the surface area of each small cube is 6× 12=6, the sum of the surface areas of all small cubes is 6n3, and the sum of the surface areas of all small cubes is twice the surface area of the original cube.
When n=2, the sum of the surface areas of all small cubes is twice that of the original cube; When n=3, the sum of the surface areas of all small cubes is three times that of the original cube.
9. (Jiangsu Province, 3 points in 2009) As shown in the figure, if the center line of trapezoidal ABCD is known and the area of △DEF is, then the area of trapezoidal ABCD is ▲ cm2.
Answer 16.
Trapezoidal mean value theorem of test sites
According to the analysis, it can be seen that the height of △DEF is half of the height of the trapezoid, so the product of the center line and the height can be obtained according to the area of the triangle, that is, the area of the trapezoid can be obtained:
Let the height of the trapezoid be h,
∫ef is the center line of trapezoidal ABCD, and the height of∴△ ∴△DEF is.
The area of △DEF is ∴.
∴ The area of trapezoidal ABCD is.
10. (Changzhou City, Jiangsu Province, 2065438+2 points in 00) As shown in the figure, there are 12 numbers in the circle, which are 0, 1, 2, 3, 4, …, 1 1 respectively. Electronic flea
Every time you jump, you can jump from one circle to an adjacent circle. Now, an electronic flea starts from the circle marked with the number "0" and presses it counterclockwise.
After jumping 20 10 times, it falls into the circle marked with numbers.
Answer 6.
Classification and induction of test sites (multiple graphics).
According to the meaning of the question, we can find that the number is 0, 1, 2, 3, 4, …, 1 1, that is, 12 is a cycle:
If the remainder is 0, the number marked by the circle is 0;
If the remainder is 1, the number of circle marks is11;
If the remainder is 2, the number marked by the circle is10;
If the remainder is 3, the number marked by the circle is 9;
…;
If the remainder is 1 1, the number of circle marks is 1.
∫20 10 divided by 12, the remainder is 6. ∴: The number marked in the circle is 6.
11.(2011Changzhou, Jiangsu, 2 points) Divide a cube with a side length of 4 into 29 cubes with an integer side length (and no remainder), in which
The number of cubes with a side length of 1 is ▲.
Answer 24.
Splicing of test center graphics.
Analysis (idea 1) The volume with side length of 4 is 64, the volume with side length of 3 is 27, the volume with side length of 2 is 8, and the volume with side length of 1 is 1.
The volumes of 29 cubes are 1, 1, 1, ... respectively 1, (1+7). ......
A ***29, the total number of volumes is 64. If 29 is removed from 1, the excess volume of 64-29 =35, and the combination with side length of 2 or 3 should be given respectively.
(1) If only the side length is 2, the extra volume is 35=7+7+7+7+7, which means it can only be 5 sides of 2/24 sides of kloc-0/.
(2) If the side length is 3, the extra volume is 35-26=9, which is not divisible and has no solution.
So there is only one possibility, 24 with a side length of 1 and 5 with a side length of 2.
(Idea 2) Case 1: Let the number of cubes with a side length of 3 be and the number of cubes with a side length of 2 be, then the number of cubes with a side length of 1 be. According to the meaning of the question
So there is no positive integer.
Case 2: If the number of cubes with side length 3 is 0 and the number of cubes with side length 1 is 0, then the number of cubes with side length 2 is 0. According to the meaning of the question.
Case 3: If the number of cubes with side length 2 is 0 and the number of cubes with side length 1 is 0, then the number of cubes with side length 3 is 0. Is there an integer solution according to the meaning of the question?
12.(20 12 Changzhou, Jiangsu, 2 points) As shown in the figure, the sum of inverse proportional functions is known. Point A is on the positive semi-axis of the Y axis, and the point A is the straight line BC∑x axis, which intersects with the images of two inverse proportional functions of point B and point C respectively, connecting OC and OB. If the area of △BOC is, AC: AB = 2: 3, then = ▲, = ▲.