The composition of the bridge No, I want the first question of thinking, investigation and Research on page 121 of pep

The composition of the bridge No, I want the first question of thinking, investigation and Research on page 121 of pep

Bridge, we are familiar with things, in the mountain stream there is a single wooden bridge, in the highway there is a stone arch bridge, in the city there are overpasses and flyovers. Bridges all over the country, each bridge is different, each bridge has a different story. It is a channel for people to survive, like a rainbow in the distance, now in the heart of the eager

English unit 3 words
In addition to words, it's better to have notes,

Babysit temporary care v. camp camping v. plan v. Tibet hike v. Hong Kong Year adv.get Send v. Postcard Postcard n. sa

Unit3 (1) answer
The answer is needed today

They are going to the zoo. 3. She's listening to the music 4. She is singing. 5. They are going camping. 6. He's playing computer games
There is no answer to the following question

1. A elevator rises from a 42 m deep mine bottom at a steady speed, then rises at a steady speed for 10 s after 5 s, and then rises at a steady speed for 3 s until the mine mouth is at a steady state, so as to find the maximum speed in the process of its movement
2. When an object is accelerating uniformly, the initial velocity is V0 and the final velocity is vt. what is the velocity when it passes through the first 2 / 3 displacement? What is the velocity when it passes through the first 2 / 3 time?
3. For a car that starts to accelerate uniformly from a standstill, the distance between the starting point and a point is 125m, which lasts for 10s, and the speed when passing B point is 15m / s

Analysis 1: the physical process in the first question is to carry out uniform acceleration motion to increase its speed, then increase its speed at a uniform speed, and then decrease its speed. From this, we can know that the speed of the uniform process is the largest in the rising process
1: It is easy to know from the question that the speed of rising at constant speed is the maximum, so let the maximum speed be v. the acceleration of accelerating rising is A1; the deceleration of decelerating rising is A2; the acceleration of accelerating rising is A1; the deceleration of decelerating rising is A2;
V = at; s = 1 / 2 * a * t * t; therefore, the following three equations are established: v = 5A1; v = 3a2; 42 = 1 / 2 * A1 * 25 + 10 * V + 1 / 2 * A2 * 9
The solution of these three equations is v = 3m / s
Analysis 2: this problem is to examine our flexible use of "2As = VT * vt-v0 * V0"
2:2as=Vt*Vt-V0*V0
Therefore, several equations are listed from this formula; the velocity of 2 / 3 displacement is calculated as: root sign (1 / 3 * V0 * V0 + 2 / 3vt * VT)
Vt=V0+at;
Therefore, several equations are listed to calculate the velocity of two-thirds of the time: (1 / 3v0 + 2 / 3vt)
Analysis 3: first of all, we need to make clear the physical process of this problem. The car has been doing uniform acceleration, and there will be a speed at point A. We can set it, and there will also be a speed at point B, which can be expressed by the speed of point a, that is: VB = VA + 10A; the time from a to B is 10 seconds, and the speed of point B is 15 meters per second, so it is easy to get the following results:
3: 2As = VB * VB VA * VA; VB = VA + at; VA = 10m / S; a = 0.5m / s * s
So the time to point a is VA = at; so t = 20s; then s = 100m

Volume, bottom area, perimeter, side area of cylinder How to ask

Volume: bottom area times height v = sh
Bottom area: π times the square of the bottom radius s = π R2
Perimeter of bottom surface: π times diameter of bottom surface C = π d = 2 π R
Side area: bottom perimeter times height s = Ch

84 4 / 119 * 1.375 + 105 5 / 119 * 0.9=
84 4 / 119 * 1.375 + 105 5 / 119 * 0.9=

Original formula = (84 + 4 / 119) * 1.375 + (105 + 5 / 119) * 0.9 = 84 * ` 1.375 + 105 * 0.9 + (4 * 1.375 + 4.5) / 119
=84*(1+3/8)+94.5+9/119
=84+21*3/2+94.5+9/119
=210 9 / 119

If a is the third quadrant, simplify cosa * (1 / Sina) * √ (1 / cos & sup2; a) - 1

A is the third quadrant, so cosa0,) √ (1 / cos & sup2; a) = - 1 / cosa, then the original formula becomes - (1 / Sina) - 1 = - csca-1

As shown in the figure, a road with the same width is built on the rectangular ground with a width of 20 meters and a length of 32 meters (the shaded part in the figure), and the rest is planted with lawn. To make the lawn area 540 square meters, the width of the road is ()
A. 5m B. 4m C. 3M D. 2m

Let the width of the road be X. according to the meaning of the title, we get 20x + 32x-x2 = 20 × 32-540, sort out (x-26) 2 = 576, square root x-26 = 24 or x-26 = - 24, and solve x = 50 (rounding off) or x = 2, so the width of the road is 2m

How to find the ratio of 3 / 8 to 9 / 16?
How to find the ratio of 3 / 8 to 9 / 16? 39:13 of 14? 25 kg: half a ton?
.........................................................................................

3 out of 8: 9 out of 16
=3 / 8 x 16 / 9
=2 / 3
Dividing by a fraction is equal to multiplying by the reciprocal of the fraction. How about multiplying fractions?

An is an arithmetic sequence, the first term A1 > 0 a2007 + A2008 > 0 a2007 × a20080 s4015

a(n)=a+(n-1)d,
a(1)=a>0,
If d > = 0, then a (n) > 0, a (2007) * a (2008) > 0, and a (2007) * a (2008)