Answer:

Step-by-step explanation:
![(x+1-i)(x+1+i)\\\\=(x+1)^2 -i^2~~~~~~~~~~~~~;[a^2 -b^2 = (a+b)(a-b)]\\\\=x^2 +2x +1 -(-1)\\\\=x^2 +2x +1+1\\\\=x^2 +2x +2](https://tex.z-dn.net/?f=%28x%2B1-i%29%28x%2B1%2Bi%29%5C%5C%5C%5C%3D%28x%2B1%29%5E2%20-i%5E2~~~~~~~~~~~~~%3B%5Ba%5E2%20-b%5E2%20%3D%20%28a%2Bb%29%28a-b%29%5D%5C%5C%5C%5C%3Dx%5E2%20%2B2x%20%2B1%20-%28-1%29%5C%5C%5C%5C%3Dx%5E2%20%2B2x%20%2B1%2B1%5C%5C%5C%5C%3Dx%5E2%20%2B2x%20%2B2)
Answer:

Step-by-step explanation:
<u>Slope-intercept </u><u>form</u>
y= mx +c, where m is the slope and c is the y-intercept
Line p: y= -8x +6
slope= -8
The product of the slopes of perpendicular lines is -1. Let the slope of line q be m.
m(-8)= -1
m= -1 ÷(-8)
m= ⅛
Substitute m= ⅛ into the equation:
y= ⅛x +c
To find the value of c, substitute a pair of coordinates that the line passes through into the equation.
When x= 2, y= -2,
-2= ⅛(2) +c



Thus, the equation of line q is
.
If ~v = hv1, v2, v3i and ~w = hw1, w2, w3i are vectors and c is a scalar, then
(a) c~v = hcv1, cv2, cv3i
(b) ~v + ~w = hv1 + w1, v2 + w2, v3 + w3i
(c) ~v − ~w = hv1 − w1, v2 − w2, v3 − w3i.