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(4%20-%203)%20%60times%206%20%2B%202
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4
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2^{2}
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5 masalah yang serupa dengan:
(4%20-%203)%20%60times%206%20%2B%202
Masalah Sama dari Carian Web
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 ...
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Masalah yang serupa
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) }
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