Ⅰ 有没有什么在线做数学题的网站
中小学数学网
http://www.mathcn.com/
中国数学在线
http://www.mathfan.com/
小学数学专业网
http://www.shuxueweb.com/
延安数学教育网站
http://yamaths.diy.myrice.com/
1+e数学乐园
http://www.aoshu.com/
数学网站联盟
http://www.sxlm.net/index2.asp
中学数学教学网
http://www.rasx.net/
华师大数学网站
http://www.hsdczsx.com/article_index.asp
快乐数学
http://klsx.diy.myrice.com/
数学时空
http://www.shuxue123.com/
数学教育教学资源中心
http://www.esx.net/
数学人
http://www.mathren.com/
初中数学网
http://www.czsx.com.cn/
中国奥数网
http://www.aoshu.cn/
广州市中学数学之窗
http://maths.guangztr.e.cn/index.html
高中数学网
http://www.gzmath.com/
我形我数
http://www.wxws.cn/
数学中国
http://www.madio.net/index.html
中学数学题库
http://www.tiku.net/
数学456资源网
http://www.maths456.net/
这些有的是小学，有的是中学，有的是高中
都挺好
Ⅱ 给我一个可以搜索试题的网站,可以搜索数学, 如题,外加英语,直接搜索试题哟,除了：上学吧
网络几乎什么都有,你也可以上搜狐,我记得我高中是老师印的试题就有搜狐教育四个字.希望对你有帮助
Ⅲ 谁有英文版的数学题
Suppose you have two five-card poker hands dealt from separate decks. You are told hand A contains at least one ace. You are told hand B contains the ace of spades. Which hand is more likely to contain at least one more ace? A large tank has a steadily flowing intake and 10 outlet valves, the latter being all of the same size. With 10 outlets open, it takes two and one half hours to empty the tank; with 6 outlets open it takes five and one half hours to empty the tank. After the tank is empty and with all 10 outlets closed, how long will it take to fill the tank? 到这个网站看看很多的 http://mathproblems.info/
Ⅳ 可以有效学习初中 英语 数学 和物理的网站有什么（要精选的
首推菁优网
但是这个网站只有初中数学和物理
没有英语
虽然我在这建议
但是还是少上为妙
因为有很详细的解答
会造成你撒懒的习惯
Ⅳ 在美国读本科，想知道有没有那种能帮我解答数学作业问题的网站
在美国的话，目前quizlet和chegg的功能已经很强大了，但唯一对中国学生不方便的就是全英文界面，整体也比较偏向西方用户使用习惯，国内目前也已经出现类似的网站，专门服务中国留学生，例如ikeeper，有兴趣可以去看一下
Ⅵ 求可以免费下载初中英语和数学习题资料的网站
初中英语合集网络网盘下载
链接：https://pan..com/s/1znmI8mJTas01m1m03zCRfQ
简介：初中英语优质资料下载，包括：试题试卷、课件、教材、视频、各大名师网校合集。
Ⅶ 国内和国外的专门讨论数学竞赛题网站有哪些
英文： www.artofproblemsolving.com ，全世界最牛的数学竞赛网站。 中文： www.aoshoo.com 浙江奥数网： http://www.zjaoshu.com/maths/ （以上摘自： http://..com/question/60305263.html ） http://www.isud.com.cn/down.asp?cat_id=48&class_id=556 http://www.swxl.com.cn/match/ShowClass.asp?ClassID=206&page=2 http://www.cbe21.com/subject/maths/jsyd.php 竞赛论坛： http://bbs.tesoon.com/read.php?tid=130622 不太好找呀！
Ⅷ 请问哪里有英语出的数学题目
国外的数学题和我们的可能形式不太一样把~太多了,先发几个,你去下面的网页自己选把~
This site offers several examples where representing a number in a base different from the customary 10 may have a great advantage. For example, binary representation is instrumental in solving Nim, Scoring, Turning Turtles and other puzzles. As every one knows nowadays, this is also the system underlying the modern computer technology.
Along with the binary, the science of computers employs bases 8 and 16 for it's very easy to convert between the three while using bases 8 and 16 shortens considerably number representations.
To represent 8 first digits in the binary system we need 3 bits. Thus we have, 0=000, 1=001, 2=010, 3=011, 4=100, 5=101, 6=110, 7=111. Assume M=(2065)8. In order to obtain its binary representation, replace each of the four digits with the corresponding triple of bits: 010 000 110 101. After removing the leading zeros, binary representation is immediate: M=(10000110101)2. (For the hexadecimal system conversion is quite similar, except that now one should use 4-bit representation of numbers below 16.) This fact follows from the general conversion algorithm and the observation that 8=23 (and, of course, 16=24.) Thus it appears that the shortest way to convert numbers into the binary system is to first convert them into either octal or hexadecimal representation. Now let see how to implement the general algorithm programmatically.
For the sake of reference, representation of a number in a system with base (radix) N may only consist of digits that are less than N.
More accurately, if
(1) M = akNk+ak-1Nk-1+...+a1N1+a0
with 0 <= ai < N we have a representation of M in base N system and write
M = (akak-1...a0)N
If we rewrite (1) as
(2) M = a0+N*(a1+N*(a2+N*...))
the algorithm for obtaining coefficients ai becomes more obvious. For example, a0=M molo n and a1=(M/N) molo n, and so on.
Recursive implementation
Let's represent the algorithm mnemonically: (result is a string or character variable where I shall accumulate the digits of the result one at a time)
result = ""
if M < N, result = 'M' + result. Stop.
S = M mod N, result = 'S' + result
M = M/N
goto 2
A few words of explanation.
"" is an empty string. You may remember it's a zero element for string concatenation.
Here we check whether the conversion procere is over. It's over if M is less than N in which case M is a digit (with some qualification for N>10) and no additional action is necessary. Just prepend it in front of all other digits obtained previously. The '+' plus sign stands for the string concatenation.
If we got this far, M is not less than N. First we extract its remainder of division by N, prepend this digit to the result as described previously, and reassign M to be M/N.
This says that the whole process should be repeated starting with step 2.
I would like to have a function say called Conversion that takes two arguments M and N and returns representation of the number M in base N. The function might look like this
1 String Conversion(int M, int N) // return string, accept two integers
2 {
3 if (M < N) // see if it's time to return
4 return new String(""+M); // ""+M makes a string out of a digit
5 else // the time is not yet ripe
6 return Conversion(M/N, N) +
new String(""+(M mod N)); // continue
7 }
This is virtually a working Java function and it would look very much the same in C++ and require only a slight modification for C. As you see, at some point the function calls itself with a different first argument. One may say that the function is defined in terms of itself. Such functions are called recursive. (The best known recursive function is factorial: n!=n*(n-1)!.) The function calls (applies) itself to its arguments, and then (naturally) applies itself to its new arguments, and then ... and so on. We can be sure that the process will eventually stop because the sequence of arguments (the first ones) is decreasing. Thus sooner or later the first argument will be less than the second and the process will start emerging from the recursion, still a step at a time.
Iterative implementation
Not all programming languages (Basic is one example) allow functions to call themselves recursively. Recursive functions may also be undesirable if process interruption might be expected for whatever reason. For example, in the Tower of Hanoi puzzle, the user may want to interrupt the demonstration being eager to test his or her understanding of the solution. There are complications e to the manner in which computers execute programs when one wishes to jump out of several levels of recursive calls.
Note however that the string proced by the conversion algorithm is obtained in the wrong order: all digits are computed first and then written into the string the last digit first. Recursive implementation easily got around this difficulty. With each invocation of the Conversion function, computer creates a new environment in which passed values of M, N, and the newly computed S are stored. Completing the function call, i.e. returning from the function we find the environment as it was before the call. Recursive functions store a sequence of computations implicitly. Eliminating recursive calls implies that we must manage to store the computed digits explicitly and then retrieve them in the reversed order.
In Computer Science such a mechanism is known as LIFO - Last In First Out. It's best implemented with a stack data structure. Stack admits only two operations: push and pop. Intuitively stack can be visualized as indeed a stack of objects. Objects are stacked on top of each other so that to retrieve an object one has to remove all the objects above the needed one. Obviously the only object available for immediate removal is the top one, i.e. the one that got on the stack last.
Then iterative implementation of the Conversion function might look as the following.
1 String Conversion(int M, int N) // return string, accept two integers
2 {
3 Stack stack = new Stack(); // create a stack
4 while (M >= N) // now the repetitive loop is clearly seen
5 {
6 stack.push(M mod N); // store a digit
7 M = M/N; // find new M
8 }
9 // now it's time to collect the digits together
10 String str = new String(""+M); // create a string with a single digit M
11 while (stack.NotEmpty())
12 str = str+stack.pop() // get from the stack next digit
13 return str;
14 }
The function is by far longer than its recursive counterpart; but, as I said, sometimes it's the one you want to use, and sometimes it's the only one you may actually use.
Divisibility Criteria
Divisibility criteria are ways of telling whether one number divides another without actually carrying the division through. Implicit in this definition is the assumption that the criteria in question affords a simpler way than the straight division to answer the question of divisibility. Divisibility criteria re constructed in terms of the digits that compose a given number. To fix the notation, A will be the number whose divisibility by another number d we are going to investigate on this page. In the decimal system,
A = 10nan + 10n-1an-1 + ... + 101a1 + a0
an0. We readily have several examples:
Divisibility by 3. Let s+(A) = an + an-1 + ... + a0. Then A is divisible by 3 iff s+(A) is.
Divisibility by 9. With the same function s+, A is divisible by 9 iff s+(A) is.
Divisibility by 11. Let s±(A) = a0 - a1 + ... (-1)nan. Then A is divisible by 11 iff s±(A) is.
They all follow from the two basic properties of the molo arithmetic
[A]d + [B]d = [A + B]d
[A]d·[B]d = [A·B]d
and the fact that 10 = 1 (mod 9) and 10 = -1 (mod 11), from which we successively get 102 = 1 (mod 9) and 102 = 1 (mod 11), 103 = 1 (mod 9) and 103 = -1 (mod 11), and so on.
Note that both s+(A) and s±(A) are linear combinations of the digits of A. This is the kind of functions we shall allow on this page. (One generalization would be to consider other bases.)
We formalize the definition the following way:
Definition
A function f(A) = f(an, ..., a0) is called a divisibility criterion by an integer d provided, starting with some A, |f(A)| < A and A is divisible by d iff f(A) is divisible by d.
O(d) is defined as the set of all divisibility criteria by d. Here are a few examples:
s + O(9) and s + O(3). (Incidently, O(9)O(3). Why?)
s±O(11)
f1(A) = a0 O(2)O(5)
f2(A) = 10a1 + a0 O(4)O(25)
f3(A) = 102a2 + 10a1 + a0 O(8)O(125)
f4(A) = 2a1 + a0 O(4)
We have more. Indeed, since 100 = 1 (mod 11),
f5(A) = (a1a0)10 + (a3a2)10 + (a5a4)10 + ...O(11)
(f5 is obtained by splitting A right-to-left into 2-digit numbers.) Similarly,
f6(A) = (a1a0)10 - (a3a2)10 + (a5a4)10 - ...O(101)
In the same spirit,
f7(A) = (a2a1a0)10 - (a5a4a3)10 + (a8a7a6)10 - ...O(1001)
Interestingly, since 1001 = 7·11·13, f7O(7)O(11)O(13). The fact may appear uninspiring for it does not relieve one from drudging through the division by 7 or 11 or 13. However, in some cases this rule is of great help indeed:
2,003,008 is divisible by 7 for so is (008) - (003) + 2 = 7.
524784 is divisible by 13 for so is 784 - 524 = 260.
Would you rather go on with the long division?
Stuart Anderson developed a general framework for deriving divisibility criteria. In particular, he noticed that
f8(A) = 2·(... 2·(2·(anan-1)10 + (an-2an-3)10) + (an-4an-5)10) + ...O(7),
which holds for odd n. For even n, modification is obvious. This also follows from the fact that 102 = 2 (mod 7).
There is another approach that uses the following generalization of the Euclid's Proposition VII.30:
Let a and d be mutually prime (coprime). Then d|ab is equivalent to d|b.
Let d be a divisor of (10c - 1) for some c. Then clearly d and c are coprime. Denote
A1 = 10n-1an + 10n-2an-1 + ... + a1,
so that A = 10A1 + a0. We have
Ac = (ca0 + A1) + (10c - 1)A1
from which it follows that
f(A) = ca0 + A1O(d)
This leads to a recursive criterion. For example, let d = 19, c = 2. Then (10c - 1) is divisible by 19. Given a number A, remove a0, add 2a0 to the remaining number A1. Proceed with these steps until you obtain a number which is obviously divisible by 19 or is obviously not divisible by 19. Whatever the case, the same will be true of the original number A. Thus, we get a sequence 12311, 1233, 129, 30. The latter is not divisible by 19. Hence neither is 12311. On the other hand, as the same calculations show, 20311 is divisible by 19. (Additional examples are available elsewhere. An ingenious example was found by Gustavo Toja from Brasil.)
Ⅸ 谁能给出全英文的数学题网站
chasedream里面GMAT中有数学的题。
http://www.math.com/
Ⅹ 哪3个网站可以查询语文数学英语的问题
语文和数学: http://www.ruiwen.com/英语: http://www.4ewriting.com/