In the second year of junior high school, to find answers and solution ideas, there must be solution

Question 1 Choice B Juxtaposition is to place them side by side, regardless of priority Partial Positive The relationship between the modified and the modified The verb-object relationship is that the predicate and the object have a paired relationship Supplement Further enrich so I feel that choose B

The second question is the one that I honestly won't choose the longest, because in order to perfect the answer, it will be a little longer.

Question 3. I forgot. The worst thing I had in my second year of junior high was biology.

The fourth question must be D, this still needs to be asked.

The fifth question is B, this kind of question uses my experience to tell you that it is a good choice, and it is also necessary to detach from the material spirit.

Question 6 choose B, choose good words.

Question 7 is selected c, as above.

Question 8 choose A or D, I'm not sure, this sentence means: don't impose on others what you don't want.

I'm not sure if it's right, after all, I haven't read a book in the second year of junior high school for a long time, and we don't take this kind of question here, and the teacher doesn't talk about it in too much detail. Let's take a look at it, but we still have to consider and consider.

3. Option D idea: measures to protect susceptible populations should be measures to isolate the possibility of inducing pathogens;

4. Choose D ideas: Before making any decision, you must understand the situation clearly.

6. Choose B idea: Since it is a friend, it is a relationship that shares joys, sorrows, and sorrows together.

8. Choose D ideas: What you don't want to do doesn't mean that others want to do it Don't ask others to do what you don't want or don't want to do, it's a matter of respecting others' practices.

According to the monitoring of the Hainan Provincial Meteorological Observatory, this year's No. 11 typhoon Ute was located in the eastern part of the South China Sea, about 920 kilometers southeast of Wenchang, Hainan at 3 p.m. on August 12.

Proof is that d is dn ac, the vertical foot is n, and if db and dc are connected, dn = df (the bisector property of the remainder guess angle), db = dc (the perpendicular bisector property of the line segment) and df ab, dn ac, dfb = dnc = 90°, in the destruction of rt dbf and vertical rt dcn.

db=dc,df=dn

rt△dbf≌rt△dcn（

BF = CN in RT DFA and RT DNA.

dn=df, da=da(common edge).

rt△dfa≌rt△dna（

an=afbf=cn=ac+an=ac+af, i.e., bf=ac+af

Connect bf, because the triangle is an equilateral triangle and ad is the midline, so ad bc and bf = cf

This translates the problem into a minimum of BF+EF.

According to the fact that the sum of the two sides of the triangle is greater than the third side, it is obvious that the minimum value of bf+ef is the length e of be and the midpoint of ac, so be = 2 times the root number 3

Solution: The useful work done by tensile force w=gh=700N, =w has w total, w total = w has =140J80%=175J

Useful power 700 joules.

Total work 140 80% 175 joules.

Don't be fooled by a rally of 250N.

Don't you have a picture? W has =gh=700N*

n=w has w total*100%, w total=w has n=140j

W total = FS = 175J

Mechanical efficiency = w useful w total = mgh w pull = 80%.

The solution is wla = 175j

Can't help you. I won't.

Originally, it was 275 together, so the slow train drove for an hour means that it drove 50km, so the distance at this time was only 275-50=225.

Now the two cars are driving together, and the speeds should be added up, that is, the speed at which they are approaching is (50+75) km h

Then the distance 225 divided by the speed 125 is the time when they meet, 225 125 =, this is calculated from the start of the express train.

Then for the slow train, you have to add the first hour, that is, the answer is to meet the fast train after starting from the slow train.

Set the express train departure xh, that is, the slow train departure x+1h

50（x+1）+75x=275

x=So the slow train set off and the two cars met.

Answer: Bai1, Da, and EA are DU, respectively

The bisector of bac and baf, cad= bad, bae= fae, and dao caf is an inner flat angle, bad+ bae=90°, dae capacity 90°, da is isosceles, da is the apex bisector, ad bc trilinear is a certain reason, quadrilateral daeb is a rectangle, ab=de rectangle diagonal is equal.

Proof: Take the midpoint E of BC and connect AE. Then there is ac=ec, ae=be, so aec= cae, abe= eab, so, aec= b+ eab=2 b

And because c=2 b

So aec= cae c

So c=60°, b=30°

So a=90°