Tianjin grade two English Volume II 119 text translation?

Tianjin grade two English Volume II 119 text translation?

I forgot to bring my book back, or I can help you

English translation
module1 unit2 wonders of the world
module2 unit2 great books
module3 unit2 sporting life
Unit 2 text translation is not a dialogue

Give me 20 points and I'll call you immediately. Module1 is the greatest wonder of nature. When I arrived, it was still early in the morning and it was still raining. I got out of the car, went through the gate and walked along a path. In the East, the sky became white, but on the side of the road, it was still dark. I knew where it was, but I couldn't see anything

Given the function f (x) = 4 ^ X / (4 ^ x + 2) (1) find the value of F (a) + F (1-A) (2) find f (1 / 100) + F (2 / 100) + F (3 / 100) + +f(99/100）

f(x)=4^x/(4^x+2）=1-2/(4^x+2）f(a)+f(1-a)=1-2/(4^a+2）+1-2/(4^(1-a)+2）=2-2[1/(4^a+2）+1/(4/4^a+2)]=2-2[1/(4^a+2）+4^a/(2*4^a+4)]=2-2[1/(4^a+2）+1/2*4^a/(4^a+2)]=2-2(1+1/2*4^a)/(4^a+2）=1f(1/100)+f(2/...

Father and son go to the park for morning exercise in the morning. It takes 30 minutes for father to run from home to the park and 20 minutes for son. If father starts 5 minutes earlier than son, it takes ()
A. 8 minutes B. 9 minutes C. 10 minutes D. 11 minutes

Suppose that it takes X minutes for a son to catch up with his father

sin43°cos13°

sin43°cos13°-cos43°sin13°?

The velocity expression of a particle moving in a straight line with uniform velocity is v = (5-4t) m / s,
When t = s, the velocity of the object is 0? What is the displacement when the particle velocity is 0?

The velocity at the end of 4S is v = (5-4 * 4) - 11m / s
Let v = (5-4t) m / S = 0, then t = 1.25s, that is, when t = 1.25s, the velocity of the object is 0
a=(-11-0)/(4-1.25)=-4m/s^2
When the particle velocity is 0, the displacement is s = v0t + 1 / 2at ^ 2 = 5 * 1.25 + 1 / 2 * (- 4) * 1.25 ^ 2 = 3.125m

3 (5x + 4) = 8x + 96,6 (x-1.5) divided by 3 = x + 0.6,5 (x + 3) = 4 (x = 11)

3（5X+4)=8X+96
15X+12=8X+96
15X-8X=96-12
7X=84
X=12
6(X-1.5)/3=X+0.6
6(X-1.5)=3(X+0.6)
6X-9=3X+1.8
6X-3X=1.8+9
3X=10.8
X=3.6
5(X+3)=4(X+11)
5X+15=4X+44
5X-4X=44-15
X=29

If X & # 179; + X & # 178; + X + 1 = 0, then the power of X is 2012=

x³﹢x²﹢x＋1＝0
The factorization results in (X & # 178; + 1) (x + 1) = 0
So x + 1 = 0, x = - 1
x^2012=（-1）^2012=1

The known parabola y = - X & # 178; + 4x-3
1. Draw a sketch of parabola. 2. Observe your sketch and write down the value range of X. when the function value y is less than 0

Given △ ABC, a (2a, B-3) B (- 2,4) C (- 1,3), translate △ ABC down 6 units to get △ a'b'c ', and then translate it right 5 units to get △ a2b2c2, if A2 coordinate is (1 / 2b-2, A-1)
(1) : find the value of a and B

A (2a, B-3) finally becomes A2 (2a + 5, b-3-6) by translation
∴2a+5=1/2b-2 b-3-6=a-1
That is, 4a-b = - 14, A-B = - 8
By subtracting the two equations, 3A = - 6, a = - 2
2-B = - 8, B = 6
∴a=-2 b=6