Perimeter of rectangle=66
length of rectangle=L
width of rectangle=w
P of a rect.= 2(length)+ 2(width)
66= 2L+2w
if the length is 7in more than the width, then
L=7+w
Now we will substitute 7+w in for L. Here is our new equation:
66=2(7+w) + 2w
Solve for w
66=14+2w+2w
66=14+4w
52=4w
w=13
L=7+13, so L=20
I hooe this is explained well enough
Solution:
Given: zj = 3- j2, zz = - 4 + j3
Part a:
zj = 3- j2 = 
zz = - 4 + j3 = 
Part b:
|zj| = |3− j2| = √3² +(−2)² = √13 = 3.60
|z1| = √(3− j2)(3+ j2) = √
13 = 3.60.
Part c:
With the help of part a:
z1z2 = 