We need answers for the final English test of the fourth grade volume II of the people's education press 2011 Don't say you have to study hard, not lazy, just review, send up, thanks again

We need answers for the final English test of the fourth grade volume II of the people's education press 2011 Don't say you have to study hard, not lazy, just review, send up, thanks again

The papers are different every year ~ ~ ~ it's useless to send them to you. You'd better study hard and get a good score!

The sender of final examination paper of PEP third grade English volume 2

PEP third grade English volume 2 final paper:
Final examination paper of third grade mathematics volume II of PEP:
PEP third grade Chinese Volume II final paper:

PEP grade 4 final English paper

Oh! I've passed the exam for two years. I want to know and forget!

In the triangle ABC, de / / BC, s triangle ade: s quadrilateral bced = 1:2,

BC = 2 times root 6 is not used, because s triangle ade: s quadrilateral bced = 1:2, so s Δ ade: s Δ ABC = s Δ ade: (s Δ ade + s quadrilateral bced) = 1:3, because de / / BC so Δ ABC ∽ Δ ade so AE: AC = 1: √ 3, because AC = AE + EC so AE: EC = 1: ((√ 3) - 1) because triangle ade and triangle

F (x) = │ x ^ 2-8 │, calculate the increase and decrease interval of the function according to the derivative

When x ≥ 2 √ 2, or X ≤ - 2 √ 2, f (x) = x & # 178; - 8;
-When 2 √ 2 ≤ x ≤ 2 √ 2, f (x) = 8-x & # 178;;
So the increasing interval of F (x) is: [- 2 √ 2,0]; [2 √ 2, + ∞)
The subtraction interval is: (- ∞, - 2 √ 2]; [0,2 √ 2]

Given point a (0,2) and circle x ^ 2 + y ^ 2 = 16, points B and C are two moving points on the circle. If Ba is perpendicular to Ca, find the trajectory equation of point m in BC and explain what curve it is
Oh, I see

Let m (x, y) be the midpoint of BC
Connect OM, am, ob
In RT triangle ABC, am = MB (am is the middle line of triangle)
R ^ 2 = MB ^ 2 + om ^ 2 in RT triangle OBM
16=x^2+(y-2)^2+x^2+y^2
Well organized
x^2+y^2-2*y-6=0
It's a circle

As shown in the figure, the equilateral triangle abd and the equilateral triangle ace, try to explain be = CD

Because △ abd and △ ace are equilateral triangles, so Da = Ba, EA = ca. because △ abd and △ ace are equilateral triangles, so ∠ bad = 60 degrees, ∠ EAC = 60 degrees. So ∠ bad + ∠ DAE = ∠ EAC + ∠ DAE, so, ∠ BAE = ∠ DAC. So, in △ abd and △ ace, Ba = Da, ∠ BAE = ∠ DAC, EA = ca. so, △ abd ≌ ace. So be = CD (SAS)

If 1

Let a = (x, 2), B = (x + N, 2x-3 / 2), and the sum of the minimum and maximum values of the function FX = AB on [0,1] be an
The sum of the first n terms of sequence BN satisfies Sn + 4bn = n (n ∈ n *)
Ask an,
2: It is proved that {bn-1} is equal ratio, and that BN BN = - (4 / 5) ∧ (n + 1) + 1 is right?
3: Let CN = - an (bn-1), in the sequence CN, is there a positive integer k such that for any positive integer n, CN ≤ CK?
#Open sesame#

1: F (x) = x ^ 2 + (n + 4) x-3 it is easy to know that f (x) is simple increasing on the interval, an = f (0) + F (1) = n-1, n belongs to n * 2: when n > = 2, Sn = n-4bn; sn-1 = n-1-4 (bn-1) BN = 1 + 4 (bn-1) - 4 (BN) 5 (BN) - 5 = 4 (bn-1) - 4 (BN) - 1 = 4 / 5 * [(bn-1) - 1] so {bn-1} is GP, when n = 1, 5B1 = 1, B1 = 1 / 5

Additional questions: as shown in the figure, it is known that △ ABC is inscribed in ⊙ o, AB is the diameter, ∠ CAE = ∠ B. verification: AE and ⊙ o are tangent to point a

It is proved that ∵ AB is the diameter, ∵ ACB = 90 degree, ∵ BAC + ∵ B = 90 degree, and ∵ CAE = ∵ B, ∵ BAC + ∵ CAE = 90 degree, that is, ∵ BAE = 90 degree, so AE and ⊙ o are tangent to point a