6(2k+1)=42
(1)/( x2x^{2}x2+2)=0
(1)/( x2x^{2}x2+1)=0
4p-3=13(p=-4)
(1)/( x2x^{2}x2+1)
6(2k+1)=
- 10+{ 3-[ 2 - 15+ 9+( 2- 7)- 6]}+4
Calcule
∥3⃗u−5⃗v+ ⃗w∥
dado ⃗u =⟨2, −2, 3⟩, ⃗v =⟨1, −3, 4⟩, ⃗w=⟨3, 6,−4⟩:
0+q+e+2+a=
(150-5) - [14+89-6+3)]
