Is there anyone who is good at math junior high school, solve problems with me?

1.If the bidding with the bottom of the bid is adopted, the tenderer believes that the bottom of the bid is valid, and the bidder should float down on the basis of the bottom of the bid, and the bidder ** exceeds the bottom of the bid (93 96) and considers the bid to be invalid.

3.There is a bid bottom bidding, which can effectively reduce the bidder's bid-rigging and lead to the situation of elevation, but because of the bid bottom, it will also lead to a lot less preferential treatment for the bidder. For bidding without a bid floor, the bidder may raise the grade, which will directly lead to the loss of the tenderer, and of course, it may also be that the bidder's ** is low and the tenderer will benefit.

4.Both methods have their advantages and disadvantages, the current bidding usually adopts the bidding with the bottom of the bid, requiring the bidder to bid not more than a certain **, otherwise it is considered to be scrapped.

I scored 100-115 in junior high school math, do you think it's good? I am currently in my first year of high school.

At? I still know people in junior high school, and I can ask about high school.

In my second year of junior high school, I took the school exam 5 times, 2 times with perfect scores, and 3 times with 118 questions.

Let's talk about the topic!

The first equation first calculates the numerator directly, then the denominator is squared, the second equation is directly squared, and finally the numerator outside the parentheses is calculated first and then the numerator and denominator is multiplied by the denominator.

Analysis: 2 meters seconds = 120 meters minutes, 5 meters seconds = 300 meters minutes, when the first puppy catches up with Xiao Ming, the distance between the two puppies is 6 minutes, assuming that the first puppy does not run, it can be found that the second puppy catches up with Xiao Ming It takes 10 minutes, that is, the second puppy has to run for 10 minutes; At this time, the first puppy ran back to A for 10 minutes, which adds up to 20 minutes of travel, minus the 6 minutes that coincided in the middle, which is 14 minutes.

Solution: Set Xiao Ming to set off x minutes later, the first puppy began to chase Xiao Mingde:

The first puppy catches up with Ming and returns to place A: [120x (300-120)] 2=4x 3 minutes.

6 minutes later, another puppy also set off from A to catch up with Ming, and it took him to catch up and return to A: [120(x+6) (300-120)] 2=(4x 3+8) minutes.

So: the time interval between the two puppies returning to place A is: 6 + (4 x 3 + 8) - 4 x 3 = 14 minutes.

Mark the location B where the first dog catches up with Xiao Ming, then the time for the second dog to run to B is 6 minutes later, and then the second dog has to run a distance more than the first one, this distance is 6 minutes for Xiao Ming to run (the time for the second to run to B) plus the dog chases Xiao Ming such a pursuit stage, in 6 minutes, Xiao Ming ran 720 meters, then the second one needs to catch up with him with 720 (5-2) = 240 seconds, and then the return stage is 240 seconds to return to point B, Although the time from B to A cannot be calculated, it can be known that the second dog and the first dog run at the same speed, so their time difference has nothing to do with this period.

To sum up, only the BA segment is that the second dog ran more than the first dog, it took 240*2 seconds, plus the 360 seconds of the first run, they checked 840 seconds.

The answer is 288 seconds, and the process is self-counting.

I don't know if you have learned the four-point contour, but the problem (2) can be solved by applying the properties of the four-point continuum.

Take a look below and click to enlarge;

(1) Because there is bad=90°,ab=ad, and abe= aeb in the square abcd, so abe is an isosceles triangle, there is ab=ad=ae, it can be seen that ade is also an isosceles triangle, that is, there is 1= 2, in ade it can be calculated dae=180°-2 1, then bae=90°- dae=2 1-90°, in abe, abe= aeb=(180°- bae) 2=135°- 1, so bed= aed+ aeb= 1+(135°- 1)=135°。

Because AH bisects DAE, i.e., DAH= EAH, AD=AE, AH=AH, so DAH EAH(SAS), has DH=EH, i.e., DHE is an isosceles triangle, and has HDE= HED=90°- 2, then HEG= HGE=90°-(90°- 2)= 2, we can see that EHG is an isosceles triangle, with DH=EH=GH, and in isosceles DAE, we can know AH DE from the "three-in-one", and because FG DE, AH FG, and then AB cd shows that the quadrilateral afgh is a parallelogram with af=gh, so af=dh=gh, i.e., dg=2af.

Problem 1 bed= bea aed= abe ade, so the inner angle of the quadrilateral and 360°=90° 2 bed, the solution is bed=135°.

The second sub-question is the perpendicular line of de.

<> can sell bent mountains like this, and write in the middle only stupid for reference.

Don't be lazy, just draw a picture and you're done.

Do more exercises, and understand the wrong questions in time. Usually listen carefully to the lectures and take some notes accordingly. If you don't understand the problem, you should ask the math teacher in time, and you can read some math exercises to solve the problem when you have time.

If you have the conditions [enough time, you can afford it] can go to math cram school. ‘

Go to bed early, take care of your body, don't have fear, math is just a tool, you can control it, first of all, you have to believe in yourself.

Remove them all, and the rest is single.

40 4 = 10 people.

Number of people who can only play chess + number of people who can only play Go = 48-4-4 = 40 people.

And because"The number of people who only know chess is three times as many as the number of people who can only play Go"

So the number of people who can only play chess = 30

Only the number of people who can play Go = 10

48-4-4=40

If only x people can play Go, then 3x+x=40, x=10