Disclaimer: I wrote the math, physics, chemistry and English parts of this answer myself, and checked them with others (there will be marks), which can guarantee the average accuracy of at least 85% (except for clerical errors), and the rest are not guaranteed. . .
mathematics
Note: Tang Jiangping participated in the editing of the first volume of 397 Middle School Mathematics.
Exercise 1
First, DAABBDAD；;
Second, {0, 3, 4, 5}; (- 1, 1); 1; m & ltn； a & gt 1; 88.6; -8;
Third,
16、
(1) According to the meaning of the question, a2x-3=a 1-3x.
A>0 and a ≠ 1'
∴2x-3= 1-3x
The solution is x = 4/5;
⑵、①、0 & lta & lt 1,
Then 2x-3 >: 1-3x
Get x & gt4/5
②、a & gt 1,
2x-3 & lt； 1-3x，
Get x
17、⑴A⑵F⑶E⑷C⑸D⑹B
18, the left factorization of the original formula is: (x2-8)(x+ 1)=0.
Easy to solve, x 1=2.83, x2=-2.83, x3 =-1;
19, (1), from the meaning of the question, 15+(62- 15)e-k=52.
∴e-k=37/47,k=-㏑(37/47)=0.24
⑵、 15+(62- 15)e-kt=42
The solution is t=2.32.
20、⑴、f(x)= {- 1/2 x2+300 x-20000 0≤x≤400
60000- 100 x x & gt； celebrity
⑵、
① When 0 ≤ x ≤ 400, f(x) reaches the maximum value of 25,000 when x=300.
②X & gt； 400, and f(x) reaches the maximum value 19000 when x=40 1.
To sum up, the maximum value is 25,000, and the monthly output is 300;
2 1、
(1), let ax = t, (x>0), then x=at.
Then f (t) = a/(a2+1) × (a2t-1)/at.
=a(at-a-t)/(a2- 1)
Let t 1, t2∈R and t 1
Then f (t1)-f (T2) = a (x1-x2) (x1x 2+1)/x1x 2 (a2-1).
①0 & lt； A< is at 1, and f (t 1) < f(t2).
②a & gt； 2 o'clock, the above results are also true, so f(x) is increasing function;
(2)Yi f(x)= a(a-x-ax)/(A2- 1)=-f(x)
F (x) is odd function.
The original form of ∴ is f (1-m) < f(m2- 1).
F(x) is increasing function, and its domain is (-1, 1).
∴- 1<； 1-m & lt； 1
- 1 & lt； 1-m2 & lt； 1
1-m & lt； m2- 1
The solution is 1
physics
Note: Tang Jiangping of 397 Physics I participated in the editing.
Description of movement
1.AC，ACD，B，B，A，B；
2. Uniformly accelerated straight line, 65,125;
Third, suppose that the distance from the cliff when you honk the horn for the first time is S, and the speed of the car is V.
Then it is 340×8+8v=2s and 340×6+6v=2(s-27v-8v).
Simultaneous two equations, the solution is s= 1400m, v =10 m/s.
Second, the study of uniform linear motion.
1.ABC，D，B，D，B，C；
2.6,6m/S2 or14m/S2,20;
Third,
10、
⑴、a=x/2T2=4/2=2m/s2，
⑵、x = v0t+0.5 at 2 = 0.5×5+0.5×2×52 = 27.5m；
1 1、
(1) suppose the pilot opens the parachute after t time has passed since the fall.
Then (10t+5) × (10t-5)/(2×14.3) =125.
T=6s, so the athlete leaves the ground 0.5×10× 62+125 = 305m when getting off the plane.
(2) The time for athletes to open their umbrellas and land is (60-5)/14.3 = 3.85s..
Therefore, after the athletes get off the plane, they reach the ground after 6+3.85=9.85s;
Do it, 15, 23.3, 2.0, 1.55, c
Third, interaction.
1.ABC，A，D，BD，AC，C，CD，A；
2, 22, 2, 30, reduction, sketch, 5,11.5;
Three. 13, FN slope = g/sin45 = 50 radical 2N,
FN vertical baffle = GTAN 45 = 50n
14, let the dynamic friction coefficient be u, then 40u+60u=30.
The solution is u=0.3.
Fourth, Newton's law of motion.
1.D、D、BC、AD、C、AC、C、AC；
2. 1.35× 103，0.38，(m-m) v0/mg，(m-m) v0/m。
3. The sliding friction between the object and the board at 1 1 and 30 is GSIN 30 = g/2.
At this time, the positive pressure between the board and the object is three-thirds, so the dynamic friction coefficient between the board and the object is three-thirds.
At 60, a=F/M= three-thirds;
12. It is easy to know that the speed of an athlete when he starts to touch the net is 8m/s and the speed when he leaves the net is10m/s..
Then the catenary acceleration in 1.2s is a =18/1.2 =15m/S2.
So (f = ma+g) = 60× (15+10) =1500n.
13, and the acceleration of the ring a = v2/2x = 42/1=16m/S2.
∴f=m(a+g)=0.05×( 16+ 10)= 1.3n；
∴FN=G-F=2- 1.3=0.7N