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P4=4= k∈{0,n−1), n∈N 4×3× + bi=c+di⇔a=c∧b +x r2 2×1 equals = P4=4= k∈{0,n−1), n∈N 4×3× + bi=c+di⇔a=c∧b +x r2 2×1
wala na ako maanswer talaga hahaha sumakit ulo ko sa dami ng theory na hinanap ko. Nood doon, basa dito, tae hahaha
 
P4=4= k∈{0,n−1), n∈N 4×3× + bi=c+di⇔a=c∧b +x r2 2×1 =24= k∈{0,n−1), n∈N 4×3× + bi=c+di⇔a=c∧b +x r2 2×1
 
P4=4= k∈{0,n−1), n∈N 4×3× + bi=c+di⇔a=c∧b +x r2 2×1= 4 = k∈{0,n−1), n∈N 4×3× + bi=c+di⇔a=c∧b +x r2 2×1
 
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