Jingmen City Bureau of education Hongqi primary school fifth grade Chinese final examination paper 2011 what is the title of the composition? Urgent, the final exam tomorrow I'm poor at writing. I want to know first. I can't write when I get it. Please help me!

Jingmen City Bureau of education Hongqi primary school fifth grade Chinese final examination paper 2011 what is the title of the composition? Urgent, the final exam tomorrow I'm poor at writing. I want to know first. I can't write when I get it. Please help me!


Childhood fun. Partner. Moving thing. You make this article well, and change it according to the situation in the exam



People's education press Chinese Grade 5 Volume 2 unit 7 composition
More than 450 words


An impressive person in my five years of primary school life, there are many people who have left an impression on me, including classmates, father, mother, etc. but the one who impresses me most is my good friend --- Liu. She not only studies well, but also likes to help others. She is not tall, with a pair of big eyes embedded in her face



PEP Chinese exercise paper (Grade 5, Volume 1, unit 7)
1、 Write the words according to the meaning
It refers to dealing with things or speaking decisively without hesitation
2、 Write sentences as required (2 points)
1. Change to be sentence
The seventh company decided to move to the Dragon King Temple and assigned the task of covering the masses and the company to the sixth class
——————————————————————————————
3、 Write the synonyms of the following words (3 points)
Contain () filth () respect ()


To be firm
The seventh company decided to move to the Dragon King Temple, and the task of covering the masses and the company was given to the sixth class
Love and respect



There are seven bamboo poles in a row. The first one is one meter long, and the other one is half the length of the former one. Q: how many meters is the total length of these seven bamboo poles? (thinking time: 60 seconds)


The total length of the seven bamboo poles is s = 1 + 1 × (12) 1 + 1 × (12) 2 + 1 × (12) 3 + 1 × (12) 4 + 1 × (12) 5 + 1 × (12) 6, = 1 × [1 − (12) 7] 1 − 12, = 2 - (12) 6, = 16364m



Simplification: 81-a & sup2 / A & sup2; + 6A + 9 △ A-9 / 2A + 6 + A + 15 / A + 3


(81-a²)/(a²+6a+9)÷(a-9)/(2a+6)+(a+15)/(a+3)
=(9+a)(9-a)/(a+3)²÷(a-9)/(2a+6)+(a+15)/(a+3)
=-(9+a)(2a+6)/(a+3)²+(a+15)/(a+3)
=(-18a-54-2a²+6a+a²+18a+45)/(a+3)²
=(-a²+6a-9)/(a+3)²
=-(a-3)²/(a+3)²



One side of the triangle is 3cm longer than the second side, and this side is 4cm shorter than the third side. The perimeter of the triangle is 28cm. Find the length of the shortest side


Let the length of the second side be xcm, then the length of the first side is (x + 3) cm and the length of the third side is (x + 7) cm



Given that loga (x2 + 4) + loga (Y2 + 1) = loga5 + loga (2xy-1) (a > 0, and a ≠ 1), the value of log8yx is obtained


It is known that (x2 + 4) (Y2 + 1) = 5 (2xy-1), that is, x2y2 + x2 + 4y2 + 4 = 10xy-5, that is, (x2y2-6xy + 9) + (x2 + 4y2-4xy) = 0, that is, (xy-3) 2 + (x-2y) 2 = 0. Therefore, YX = 12. Log8yx = log812 = -13



To make a 256 liter square bottomless water tank, how much is the side length of the square bottom?


Let the base be x and Y and the height be Z
XYZ = 256 is Z = 256 / XY
Minimum total area required:
opt.min .2(yz+xz)+xy=512(x+y)/xy + xy=512/x + 512/y + xy>=3*512*512
If and only if 512 / x = 512 / y = XY, the minimum value is taken, so x = y = 64



In triangle ABC, ab = 5, AC = 3, BC = 4, PQ is parallel to ab. when the area of s triangle PQC is equal to that of s quadrilateral pabq, the length of PC is obtained


Let p be on the edge of AC,
∵ s △ PQC = s quadrilateral pabq,
That is s △ ABC = s △ PQC + s quadrilateral pabq = 2S △ PQC
∵PQ‖AB,∴△PQC∽△ABC
∴S△PQC/S△ABC=(PC/AC)^2
That is, 1 / 2 = (PC / 3) ^ 2
PC = 3 √ 2 / 2



Through the parabola y = 1 / 4 (x ^ 2), two different points a and B make the tangent of the parabola to intersect at point P, which satisfies the trajectory equation of point P obtained by PA vertical Pb (1);


4Y = (x ^ 2) set point A: (m, 0.25m ^ 2) set tangent y = KX + B of a into m, and then substitute 4Y = x ^ 2, so that the root has only one solution, and the tangent is y = MX / 2-0.25m ^ 2. Point B: (n, 0.25N ^ 2) substitute n, and then substitute 4Y = x ^ 2, so that the tangent of the root has only one solution is y = NX / 2-0.25n ^ 2, and the two lines are vertical, satisfying M / 2 * n / 2 = - 1MN = -