P prepaidwifi Eternal Poster Oct 31, 2020 #283 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 Last edited: Oct 31, 2020
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
G Girl_Aidee Honorary Poster Oct 31, 2020 #292 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 =24= k∈{0,n−1), n∈N 4×3× + bi=c+di⇔a=c∧b +x r2 2×1
G Girl_Aidee Honorary Poster Oct 31, 2020 #295 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
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
G Girl_Aidee Honorary Poster Oct 31, 2020 #298 P4=4= k∈{0,n−1), n∈N 4×3× + bi=c+di⇔a=c∧b +x r2 2×1 equals = Zero(False)
G Girl_Aidee Honorary Poster Oct 31, 2020 #299 P4=4= k∈{0,n−1), n∈N 4×3× + bi=c+di⇔a=c∧b +x r2 2×1 equals = One(True)