Find the sum of the complex numbers
(2 + 4i) + (9 +61)
A. 7+ 101
B. 11 + 21
C. 7+21
D. 11 + 101

11+10i

Step-by-step explanation: (a+ib) + ( x+ iy) = ( a+ x) +i( b+ y)

so, (,2+4i) + (9+6i) = (2+9) + (4+6)i

= 11 + 10i

i used to know how to do this in 7th but i forget easily

step-by-step explanation:

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!