Ⅰ 有沒有什麼在線做數學題的網站
中小學數學網
http://www.mathcn.com/
中國數學在線
http://www.mathfan.com/
小學數學專業網
http://www.shuxueweb.com/
延安數學教育網站
http://yamaths.diy.myrice.com/
1+e數學樂園
http://www.aoshu.com/
數學網站聯盟
http://www.sxlm.net/index2.asp
中學數學教學網
http://www.rasx.net/
華師大數學網站
http://www.hsdczsx.com/article_index.asp
快樂數學
http://klsx.diy.myrice.com/
數學時空
http://www.shuxue123.com/
數學教育教學資源中心
http://www.esx.net/
數學人
http://www.mathren.com/
初中數學網
http://www.czsx.com.cn/
中國奧數網
http://www.aoshu.cn/
廣州市中學數學之窗
http://maths.guangztr.e.cn/index.html
高中數學網
http://www.gzmath.com/
我形我數
http://www.wxws.cn/
數學中國
http://www.madio.net/index.html
中學數學題庫
http://www.tiku.net/
數學456資源網
http://www.maths456.net/
這些有的是小學，有的是中學，有的是高中
都挺好
Ⅱ 給我一個可以搜索試題的網站,可以搜索數學, 如題,外加英語,直接搜索試題喲,除了：上學吧
網路幾乎什麼都有,你也可以上搜狐,我記得我高中是老師印的試題就有搜狐教育四個字.希望對你有幫助
Ⅲ 誰有英文版的數學題
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? 到這個網站看看很多的 http://mathproblems.info/
Ⅳ 可以有效學習初中 英語 數學 和物理的網站有什麼（要精選的
首推菁優網
但是這個網站只有初中數學和物理
沒有英語
雖然我在這建議
但是還是少上為妙
因為有很詳細的解答
會造成你撒懶的習慣
Ⅳ 在美國讀本科，想知道有沒有那種能幫我解答數學作業問題的網站
在美國的話，目前quizlet和chegg的功能已經很強大了，但唯一對中國學生不方便的就是全英文界面，整體也比較偏向西方用戶使用習慣，國內目前也已經出現類似的網站，專門服務中國留學生，例如ikeeper，有興趣可以去看一下
Ⅵ 求可以免費下載初中英語和數學習題資料的網站
初中英語合集網路網盤下載
鏈接：https://pan..com/s/1znmI8mJTas01m1m03zCRfQ
簡介：初中英語優質資料下載，包括：試題試卷、課件、教材、視頻、各大名師網校合集。
Ⅶ 國內和國外的專門討論數學競賽題網站有哪些
英文： www.artofproblemsolving.com ，全世界最牛的數學競賽網站。 中文： www.aoshoo.com 浙江奧數網： 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 競賽論壇： http://bbs.tesoon.com/read.php?tid=130622 不太好找呀！
Ⅷ 請問哪裡有英語出的數學題目
國外的數學題和我們的可能形式不太一樣把~太多了,先發幾個,你去下面的網頁自己選把~
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 誰能給出全英文的數學題網站
chasedream裡面GMAT中有數學的題。
http://www.math.com/
Ⅹ 哪3個網站可以查詢語文數學英語的問題
語文和數學: http://www.ruiwen.com/英語: http://www.4ewriting.com/