High score to find VFP function detailed solution, VFP problem, solve

Some people stay up late to study and review very hard, but their grades are always not improved because there is no good review method, and their brains are not good.

Can provide you with methods on revision and study:

1. Promote memory, so that the results of learning are firmly stored in the brain so that they can be accessed at any time. Some students always complain that their memory is too bad, and they can't remember what they have learned when they should use it, and they lose confidence in learning. Some students think that they should forget what they have learned anyway, and it is useless to memorize it early, and hope for a surprise attack before the exam.

However, because there are too many things to memorize before the exam, I can't remember them, so I feel very annoyed.

2. Check and fill in the gaps. There are many factors that affect learning, and it can be difficult to ensure that all factors are in the best condition in a long learning process. Therefore, it is inevitable that there will be loopholes and deficiencies in the complete learning content.

Through review, check it out by yourself and make up for it in time. All students who are in a hurry to review. The loopholes and deficiencies in learning are filled in time, so their knowledge is always relatively complete.

3. The most important point is that you can improve your memory and comprehension to make you learn more with half the effort, which is the most critical point to improve your academic performance. I have used "Nikola Tesla Training" before, after learning, I have mastered a variety of efficient brain skills, easy self-study of various subjects, and finally admitted to the university, I hope my sharing can help you, good luck!

Ha, discriminating a quadratic equation of Delta?

d=4*4-4*3*1=4>0, so d=1

The last result is 1

Let's analyze it step by step:

1. First analyze case int(x 2)=x 2, that is, x is an even number within 10, s1=s1+x 2, there are 5 cases, and the sum is 15.

2. case mod(x,3) 0, that is, the number that can be divisible by 3 within 10, there , but 6 meets the first case, that is, when x 6, the statement after the execution of s1=s1+x 2 is executed after the endcase, so only 3,9, and the result of s2 is 1+3 4.

3. case int(x 2)<> x 2, that is, odd numbers within 10, there is , but the second case has been met, so only three numbers meet the conditions, so s3=3

The answer is rounding, round(,1) and the result is.

min is the function that takes the minimum value, and 9, and the smallest is.