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3%20%60frac%7B%203%20%20%7D%7B%207%20%20%7D
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3
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3
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Kuis
5 soal serupa dengan:
3%20%60frac%7B%203%20%20%7D%7B%207%20%20%7D
Soal yang Mirip dari Pencarian Web
Find the value of equation?
https://math.stackexchange.com/questions/117004/find-the-value-of-equation
I'm guessing that first \frac bc should be \frac ba. I don't know what A.M., G.M., or H.M. are. But as for how to come up with a solution, I can see good things will happen if you multiply both ...
What sort of particles corresponds to the (1,1/2) representation of the Lorentz group?
https://physics.stackexchange.com/q/439085
I think that my question wasn't actually well defined. For starters, when you restrict to the SO(3) subgroup of the (j_1, j_2) rep. of the Lorentz group, you get the j_1 \otimes j_2 rep of ...
What's wrong with my workings? Adding fractions question. [closed]
https://math.stackexchange.com/questions/2326102/whats-wrong-with-my-workings-adding-fractions-question
If you multiply some part by any number you also must divide by that number! So you get: \frac{\left(18\left(\frac19+\frac12\right)\right)}{18}+\frac{\left(6\left(\frac13-\frac12\right)\right)}{6} = \frac{(2+9)}{18}+\frac{(2-3)}{6}=11/18-1/6=\frac{11-3}{18}=\frac{4}{9} ...
Left and Right Hand Riemann Sum
https://math.stackexchange.com/q/2912004
You're correct in your intuition, you've just made a mistake in the indexing. If you look at the link, it says that i=1,2,...,n. So, i starts from 1, not 0. That means that your first interval ...
Prove by Induction help?
https://math.stackexchange.com/questions/1465029/prove-by-induction-help/1465048
From the inductive hypothesis we have \begin{align} \sum_{j=1}^{n+1}j(j+1)(j+2)&=\sum_{j=1}^{n}j(j+1)(j+2)+(k+1)(k+2)(k+3)\\[4pt] ...
Finding an orthonormal basis of the space of all functions in P2 that are orthogonal to f(t) = t.
https://math.stackexchange.com/q/1961518
You have already had W=\{at^2+bt+c\mid b=0,a,c\in\mathbb{R}\}=\{a\cdot t^2+c\cdot 1\mid a,c\in\mathbb{R}\}\tag{1} Note that W is a vector space. Here are the simpler related questions you ...
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3 \frac{ 3 }{ 7 }
4 \frac{ 15 }{ 32 }
1 \frac{ 1 }{ 2 } +3 \frac{ 4 }{ 5 }
1 \frac{ 1 }{ 2 } -3 \frac{ 4 }{ 5 }
1 \frac{ 1 }{ 2 } \times 3 \frac{ 4 }{ 5 }
1 \frac{ 1 }{ 2 } \div 3 \frac{ 4 }{ 5 }
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