Mathematics summer homework for junior high school urgent solution .

Question 1, let's fight it yourself.

In question 2, 7x+2y-5z+11 (ax+by+cz) is definitely also divisible by 11 (a, b, c are integers).

Does 7+11a)x+(2+11b)y+(-5+11c)z have a common factor 3x-7y+12z?

If there is, then.

7+11a) 3=(2+11b) (-7)=(-5+11c) 12=integer h

Observe the first numerator, which is divisible if a = 1, and the integer h=6 is obtained

This pushes b=-4 c=7

Result: 7x+2y-5z+11(x-4y+7z)=6(3x-7y+12z).

So 3x-7y+12z is divisible by 11.

Acute angles: 15 30 45 60 75

Right angle: 90 obtuse angle: 105 120 135

Flat angle: 180

What does the second question require to prove? .The above is done correctly ...

pcs, 5, 4pcs.

2.The guy above made it.

1. The workshop should make 100 parts, which are completed by A and B, first A does it alone for 5 days, and then B does it alone for 6 days, and it is known that A and B can do 18 parts per day, and how many can be completed per day by A alone?

Solution: Let A do it alone to complete x per day, then B does it alone to complete 18-x per day 5x+6(18-x)=100

5x+108-6x=100

108-100=-5x+6x

x=8 (pcs).

Answer: A does it alone and completes 8 per day.

2. Since the rectangular paper strip is cut and folded to form an isosceles triangle, it can be seen that 3 = 4 3 and 120° angles complement each other, so 3 = 60°.

The sum of the inner angles of the triangle = 180°, 3 + 4 + 5 = 180°, so 5 = 60° and another 2 + 120 ° = 180 ° (diagonal nature of parallel lines), 2 = 60° 1 + 5 + 2 = 180°, 1 = 60°

Question 2: Because he folded the note, 1= 5

According to the fact that the two lines are parallel and the inner wrong angles are equal, so 1= 4

So 4= 5= 3=60

1.A shopping mall held a prize sales activity, each customer can participate in a certain amount of money after spending a certain amount of money**, the award number is a four-digit number composed of six numbers from 1 to 6, and the award regulations are as follows: First prize:

The number is 6666; 2nd Prize: x333; Third prize: xx22; Fourth Prize:

The last digit of the number is an even number.

Q: The probabilities of the first, second, third and fourth prizes are (1 1296), (1 216), (1 36), and 1 3 respectively

2.Psychologists have found that under normal circumstances, students' attention changes with the change of the teacher's lecture time, the student's attention gradually increases at the beginning of the lecture, and the student's attention remains in a more ideal state for a period of time in the middle, and then the student's attention begins to be scattered

Question: (1) From the table, after the lecture starts, the students' attention starts to decline from a few minutes? Guess that attention decreases during Y vs. T

and expressed in formulas.

Let this function be: y=kt+b, and substituting the values of time and attention gives this function as y=-7t+380

2) Using the relation in question (1), find what is the value of the student's attention y when t is equal to 27 points. There is an existing math problem that needs to be explained for 20 minutes, and in order to achieve better results, the students' attention span is required to reach a minimum of 190, so whether the teacher can finish the problem when the students' attention reaches the required state is the reason.

Question supplement: Class time t (points): 0 5 10 15 20 25 30 35 40 45

Student's attention y: 100 191 240 240 240 205 170 135

2) Answer: According to y=-7t+380, it can be concluded that when t=27, y=191

Yes, as you can see from the table, the attention span from 5 minutes to 25 minutes is above 190. Therefore, the question can be completed in 20 minutes.

Analysis, (1) First Prize, 1 6*1 6*1 6*1 6=1 1296 Second Prize, 1 6*1 6*1 6=1 216 Third Prize, 1 6*1 6=1 36

The fourth prize, since four are selected from 1 to 6 to form a four-digit number, then this number is either an odd number or an even number [the odd and even method is to look at the number at the end], therefore, the probability is 1 2.

2), 20 minutes later, it starts to decline, y=240-35*(t-20) 5

240-7t+140

380-7t，（45≥t≥20）

When t=27, 191.

Students are required to pay attention to a minimum of 190 and listen to the math problem in twenty minutes.

According to the chart, it can be seen that when t=5, the student's attention is 191, and when t=27, the student's attention is also 190, when 520, summary: [As long as the teacher starts to speak after 5 minutes of class, the requirements can be satisfied. Therefore, the teacher can finish the question with the student's attention at the desired level.

The four-digit number composed of six numbers to 6 has a total of: 6*6*6*6=1296, and the probability of the first prize is: 1 1296

The probability of the second prize is: 6 1296 = 1 216

The probability of the third prize is: 1 36

The probability of the fourth prize is: 1 3

6 to the 4th power 6 to the 3rd power 6 to the 2nd power 1 2

20 minutes y=minus 7x+380 Bringing t into y=191 Yes, there are 20 minutes from the fifth minute of class to the end of the 25th minute. I hope I can help you, choose me.

Is this a question from the first year of junior high school? I'm also a first-year junior high school student, so how can I feel so difficult?

Solution: Let event p be the probability of unlocking, then the basic event space of p is "0,1,2,3,4,5,6,7,8,9".

Since only one number can open the lock, then p=1 10

Hope to be adopted, thank you! ）

The possible values of p are "0,1,2,3,4,5,6,7,8,9" which is a total of 10.

Only one digit is his last password number, so p=1 10

There are a total of 10 possibilities for the last bit, 0-9

There is only one correct number, so the probability is 1 10

The total probability is 10

Correct is 1, so the probability is 1 10

10% since you forgot the last digit on the left, that's one from 0 to 9, one tenth.