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If the walking speed is x, the fast walking speed is 3x, and the slow walking speed is 2xLet the first leg be a, then the next leg is 30-a.
From the meaning of the title: 30 x=a (3x)+(30-a) (2x)+25;0 solution: So the walking speed is between to (unit: km h).
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If the walking speed is x, the fast walking speed is 3x, and the slow walking speed is 2xThe first leg is A, and the next leg is 30-A.
From the meaning of the title: 30 x=a (3x)+(30-a) (2x)+25;A is greater than or equal to 0, less than or equal to 30
So the solution is greater than or equal to less than or equal to.
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There is no solution, there is a problem with this question.
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According to the existing conditions, it is a binary linear equation with many solutions.
Personally, I think there is a problem with this question.
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Is this the title? I didn't say anything about the answers!
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Below 2x+3x+x, what speed do you count.
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Unclear question, what exactly are you asking?
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According to the problem, you can only find one equation, but there are two unknown quantities, could it be that there is a problem with the problem? Are you sure the previous paragraph doesn't have to be the first half?
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Is it the speed of walking? If yes, set the walking speed to xkm min
Solution: 30 x 30 (3x 2x x) 25
The result should be x=1
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Wouldn't it be nice to set an equation?
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Preface: Some elementary school students do very well in math in elementary school, and when they enter junior high school, they will find that their children's math scores are not good. As a parent, you should know that this is mainly because some children will let their children be like fish in water when they adapt to the primary school environment, but when they are in junior high school, the children will change their learning environment, so the children's junior high school mathematics scores are relatively poor.
At the same time, as a parent, you should know that when your child is in primary school mathematics, the curriculum for primary school students is relatively simple. It is very easy for children to interpret math problems in elementary school, but the math culture class they learn in junior high school is completely different. <>
In life, when many parents are faced with their children's primary school years, they will find that the general children's math and Chinese scores are relatively excellent. When the children reach junior high school, many children do not adapt to a new school environment, so it will directly affect the decline of children's academic performance. As a parent, if you judge that your child's math score has declined, you should educate your child earlier, and at the same time, let your child know that learning math is very important.
Faced with the fact that when the children in the family are in junior high school, the children's math scores are getting worse and worse. Parents should know that this may be due to the fact that the child's math skills are not good when they are in primary school, so their children's math scores become worse and worse when they reach junior high school. Although some children perform very well in mathematics in primary school, parents should know that in fact, some children will find their homework in primary school very simple when they go to primary school.
However, as the grade level gradually increases, the mathematics culture class will become more and more difficult, and the child's cultural foundation can be revealed. <>
If parents judge that their child's math score is getting worse and worse, parents may choose to give appropriate consideration to improving their child's narrative math score. As a parent, when helping your child improve his math score, you can choose to buy more test papers and exercises for your child, so that your child can do some test papers when he has nothing to do at home.
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It is because of mathematical thinking, junior high school mathematics pays more attention to mathematical thinking, and if a good thinking model is not established, this situation will occur.
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Because I didn't adjust my study methods and study time well when I was in junior high school, I couldn't keep up with my studies, so my grades would be bad gradually.
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This may happen when some children have problems with their learning methods and are unable to keep up with the teacher's thinking in time.
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Summary. Hello dear! Based on the question you described, Ms. Xiaoxi's answer is as follows:
Question 7: First find x1, then x -8x+4 -4 +10=0, then (x-4) =6, then x= 6+4 or x= 6-4. then substitute to:
x1+x2)/(x1x2)=(6+4+√6-4)/(6+4)*(6-4)=2√6/-10=-√6/5.
Hello dear, please send me the question, I will answer it for you.
Okay dear, wait a minute.
Dear, which question are you asking, question 8?
Okay dear, you wait.
Hello dear! Based on the question you described, Ms. Xiaoxi's answer is as follows: Question 7:
First, find x1, then x -8x+4 -4 +10=0, then (x-4) =6, then x= 6+4 or x= 6-4. Then substitution gets: (x1+x2) (x1x2)=(6+4+ 6-4) (6+4)*(6-4)=2 6 -10=- 6 5
Hello dear! According to the question you described, Mr. Xiaoxi's answer is as follows: 8. (a-1) +a-1) = (a-1) (a-1) = a(a-1) = a -a.
Hello dear! According to the question you described, Ms. Xiaoxi's answer is as follows: 13, 4 (m+2)=*m+2) 4=1 4+(m+2) (8-4m).
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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.
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Summary. What's the question? Please send it over and see...
I can't do math problems, and it's more difficult in junior high school.
What's the question? Please send it over and see...
I'm sorry, I really can't do this question, I can't do it in a short time, in order not to delay your time. I'm really sorry.
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<>Are all the children so good now?
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This problem involves the calculation of sector and bow areas, and is best solved using analytic geometry. For example, the boundary is composed of two arcs of different radii, the inner boundary is set to arc 1, and the outer part is set to arc 2, and the bow formed by arc 2 minus the bow formed by arc 1 is half of the shadow area.
Let the lower right corner of the square be the origin (0,0), and pass through both sides of the origin point that is the x-axis and y-axis; Then the equation of the circle where arc 2 is located: (x+5) 2+(y-5) 2=25x 2+y 2+10x-10y+25=0, let the circle equation where arc 1 is located: x 2+y 2=100, x 2+y 2-100=0, let the circle circle - circle get the equation where the chord is:
10x-10y+125=0,2x-2y+25=0。The centroid distance of the chord is: d =|25|/√4+4)=25√2/4;Chord length = 2 (100-1250 16) = 5 14 2;chord center area = 125 7 4;Cosine cos = (200-350 4) 200 = 9 16, the center angle is about 34°, the sector area is 100 34 360 85 9, bow = sector - chord = 85 9-125 7 4;In the same way, the bow can also be calculated; Shadow area = 2 (bow - bow).
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Such a kind of problem can use the characteristics of the graph title to carry out some rotation, which cannot be turned into a relatively simple graph algorithm, of course, there is also a general algorithm, that is, the calculation of the isosceles triangle and so on that transforms the successful deed of good deeds, and the area of these figures can be obtained by using the combination of addition and subtraction between their areas.
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1. Greater than. 2、x+y=1
Process: Absolute value of x+3 + absolute value of y-4 = 0
So x+3=0 y-4=0
x=-3 y=4
So x+y=1
3、x=0,±1,±2,±3
Process: |x|<π
then the integer x=0, 1, 2, 3
4. The absolute value of x + the absolute value of y = 3
Idea: Two absolute values are opposite to each other, indicating that both sides are 0.
i.e. x=-1, y=2
So the absolute value of x + the absolute value of y = 3
5. The minimum absolute value of the equation x+1 is 0, in this case, the value of x is -16, and the answer is -3 2
Process: Because (x-1) 0, y+1 2 0 is constant.
And the square of the absolute value of x-1 + y + half = 0x-1 = 0, y + 1 2 = 0
x=1,y=-1/2
y-x=-1/2-1=-3/2
7. Solution: when x 5, 3-x 0, x-1 0, |3-x|=x-3,|x-1|=x-1
3-x|+|x-1|=x-3+x-1=2x-48, a=-8 b=2, or a=-8 b=-2 process: absolute value of (a-b) = b-a
Therefore, the absolute value of a-b<0 a and a = 8, and the absolute value of b = 2
Therefore a=-8 b=2 or a=-8 b=-2
9. Remove the absolute value symbol.
Original = [(1 99)-(1 101)]-1 99)-(1 100)]-1 100)-(1 101)].
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Greater than 3-3, -2, -1, 0, 1, 2, 3 -1 is -8, b is -2
9.minus one hundred thousand two thousand one hundred and two.
That's pretty much it, pick me!
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First read more examples, and then combine them to learn, and you will find that it is not that difficult!!
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Oh, junior high school mathematics is fundamentally different from primary school, primary school is several classes are repeated to talk about a knowledge point, a class is divided into several sessions to on, and junior high school is almost a few classes combined into a class to on, the amount of knowledge in a class is very large, now you are only the first year of junior high school, it is simple, but you must listen carefully, do not listen to the front, it will be like you now, class listening, but when doing homework and exams, it will not, which means that you do not grasp it enough, and some problems combined with the previous content can not be done.
If you can, it is recommended that you can find your math teacher to make up for it, and then you can also learn some of the content in the back first, if the teacher does not tutor, it is best to find someone who has teaching experience and has corrected the high school examination papers, otherwise it will be a waste of money and a waste of time.
English is still quite simple, if you have learned in elementary school, it is relatively easy, every day to memorize the words of the text taught by the teacher, every time you memorize the words of the day, and then look back at the words you memorized before, continue to consolidate, and you can memorize it in a month.
As a junior high school senior, I still want to say that mathematics must lay a good foundation, and it will be much easier to learn later, otherwise I can't understand even if I want to listen to it later, and in junior high school, many people with poor grades in primary school listen carefully, and their grades will go up soon, and they will regress otherwise.
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Read more, memorize more, and preview before class.
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This may be a problem that many people of this age will be troubled by, and worrying about learning shows that the landlord wants to learn well, which is rare. But if you get too much into it, you will have a lot of troubles.
As a senior sister, I will sort out my confusion with you. When it comes to mathematics in junior high school, you may have a lot of twists and turns at this stage, so I find it difficult, and I can only get 52 points in mathematics at that time, and I don't know how to do it at all, but I can't figure out the clue of solving the problem. But in the third year of junior high school, I reintegrated all the knowledge points from beginning to end with my own understanding, and if I didn't understand, I went to the teacher.
To put it simply, one day I suddenly became familiar with mathematics, and then mathematics was my favorite subject in the high school entrance examination. It's strange that after leaving the things you bother with for a while, it is inexplicably simple on its own. It is estimated that the landlord will also have such a day.
English, as in the case of Chinese, after the landlord remembers words and sentence patterns, he can get good grades by understanding.
Remember that learning with friends and teachers is the fastest way to do this, and you must listen attentively during class. Heart, the word is very simple, and not many people can really do it.
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Mathematics is to master the ideas and formulas of problem solving, and English can only be memorized, listened more, and practiced more.
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Talk to the teacher and find out why.
Maybe there's a problem with the difficulty, the elementary school is too easy.
Practice more math, let the teacher give a few more questions every day (a little easier), and then slowly increase the amount and difficulty of the questions, or find the problem to do by yourself, find a workbook to do.
In English, you should memorize more, memorize more, usage, words and so on.
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Buy this textbook to solve it all, look at the example questions above, do the following questions, ask the teacher if you don't know, there will be progress, if you really can't think of asking, summarize the method and remember the type of questions you have done.
exists, shifts the term to obtain: -m-2>(3-m)x, and it is easy to know that if m exists, the system of equations: >>>More
Assuming that the speed of A and B is that the distance between the two stations is Z, the first time is after the t1 time, and the second time is T2, when the first encounter, A and B have traveled the whole distance Z, because the first time they meet 12 kilometers west of the midpoint, and it is 12kmt1=z 2-12 xThe second meeting place is 20 kilometers away from the East Railway Station, and A has already traveled Z+20kmT2=Z+20 x The system of equations (Z 2-12) (x+y) x=Z can be obtained from the above; (x+y)(z+20) x=3zEquation2 Equation 1 yields (2z+40) (z-24)=3 and z=112
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384 kg I calculated that five eighths of 6 baskets, three-eighths of some baskets and 24 kg more, so it was calculated that the integer baskets it contained were 3, so that each basket could hold 40 kg, so there were 384 kg in total.