Let z1 = a1 + b1i, z2 = a2 + b2i, and z3 = a3 + b3i. prove the folowing using algebra or by showing with vectors.
a. z1 + z2 = z2 + z1
b. z1 + (z2 + z3) = (z1 + z2) + z3

a)z1 +z2 =z2 + z1 ...proved.

b) z1 + ( z2+ z3 )=(z1+z2)+z3 ... proved.

Step-by-step explanation:

It is given that there are three vectors z1 = a1 + ib1, z2 = a2 + ib2 and z3 = a3 + ib3. Now, we have to prove (a) z1 + z2 = z2 + z1 and (b) z1 + (z2 +z3) = (z1 + z2) + z3.

(a) z1 + z2 = (a1 +ib1) + (a2+ ib2) = (a1 +a2) + i(b1 +b2) {Adding the real and imaginary parts separately}

Again, z2 + z1 =(a2 +ib2) + (a1 +ib1) = (a2 +a1) + i(b2 +b1) {Adding the real and imaginary parts separately}

Hence, z1 +z2 =z2 + z1 {Since, (a1 +a2) = (a2 +a1) and (b1 +b2) = (b2 +b1)}

(b) z1 + ( z2+ z3 ) = [a1 + ib1] + [(a2 + a3 ) + i(b2 + b3 )] = ( a1 + a2 + a3) + i( b1+ b2+b3) {Adding the real and imaginary parts separately}

Again, (z1+z2)+z3 = [(a1+a2) +i(b1+b2)]+[a3+ib3] = ( a1 + a2 + a3) + i( b1+ b2+b3) {Adding the real and imaginary parts separately}

Hence, z1 + ( z2+ z3 )=(z1+z2)+z3 proved.

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!