跳到主要內容
Microsoft
|
Math Solver
解決
玩
練習
下載
解決
練習
玩
遊戲中心
樂趣 + 提高技能 = 贏!
主題
代數前
意味 著
模式
最大的共同因素
最小公共倍數
動作順序
分數
混合分數
優質保理
指數
基
代數
組合類似條款
變數的求解
因素
擴大
評估分數
線性方程
二次方程
不等式
方程式系統
矩陣
三角
簡化
評價
圖
求解方程
微積分
衍生物
積分
限制
代數計算機
三角計算機
微積分計算機
矩陣計算機
下載
遊戲中心
樂趣 + 提高技能 = 贏!
主題
代數前
意味 著
模式
最大的共同因素
最小公共倍數
動作順序
分數
混合分數
優質保理
指數
基
代數
組合類似條款
變數的求解
因素
擴大
評估分數
線性方程
二次方程
不等式
方程式系統
矩陣
三角
簡化
評價
圖
求解方程
微積分
衍生物
積分
限制
代數計算機
三角計算機
微積分計算機
矩陣計算機
解決
代數
三角
統計
微積分
矩陣
變數
清單
(4%20-%203)%20%60times%206%20%2B%202
評估
4
因式分解
2^{2}
圖表
測驗
(4%20-%203)%20%60times%206%20%2B%202
來自 Web 搜索的類似問題
How do you write \displaystyle{40}\times{39}\times{38}\times{9}\times{8}\times{7} as a ratio of factorials?
https://socratic.org/questions/how-do-you-write-40times39times38times9times8times7-as-a-ratio-of-factorials
\displaystyle\frac{{{40}!\times{9}!}}{{{37}!\times{6}!}} Explanation: \displaystyle\frac{{{40}!\times{9}!}}{{{37}!\times{6}!}}
From unary to binary numeral system
https://math.stackexchange.com/q/2629441
Your revised axiom system allows you to prove all true equalities between (variable-free) terms in your language. Namely, it proves that 1 followed by any sequence of operators is equal to 1 ...
Find a divisor satisfying a given congruence
https://math.stackexchange.com/questions/367350/find-a-divisor-satisfying-a-given-congruence
Coppersmith et al. give an algorithm based on LLL lattice reduction in a paper Divisors in Residue Classes, Constructively (2004) that solves this problem if M is not too small relative to N, ...
Probability of System Working
https://math.stackexchange.com/q/2896273
For the system to work, we certainly need the micro controller to be functioning, the probability of that would be 0.8. We also need at least two peripheral devices to work. \binom{3}{2}(0.7^2)(0.3)+(0.7)^3 ...
Compute discrete logarithm
https://math.stackexchange.com/questions/557411/compute-discrete-logarithm/561569
I guess that you are supposed to take advantage of the fact that p-1=2^{19}\cdot41. With nothing else to go by, let's see where that takes us. Is 7 a primitive root? A round of repeated squaring ...
Solving permutation problem the harder way
https://math.stackexchange.com/questions/470382/solving-permutation-problem-the-harder-way
Your solution is fine, except in Case 4. In that case, there are only half as many ways, since the two groups of two girls together can be interchanged. You might have a simpler way to count this ...
更多結果
共享
復制
已復制到剪貼板
類似問題
4 - 3 \times 6 + 2
(4 - 3) \times 6 + 2
4 - 3 \times (6 + 2) ^ 2
\frac{4-3}{6}+2^2
5-4(7-9(5-1)) \times 3^3 -4
12-2(7-4)^2 \div 4
\frac{ \left( 4-3 \right) + { \left( 1+2 \right) }^{ 2 } }{ 6+ \left( 7-5 \right) }
返回頂部