A plane parallel to vectors <9, 6, 2> and <-8, -4, -5> is given by

Since the plane passes through point (5, -5, 8), then
-22(5) + 29(-5) + 12(8) = -110 - 145 + 96 = -159
The required plane is -22x + 29y + 12z = -159
z = 11/6x - 29/12y - 53/4
Answer:
Options A, D and F
Step-by-step explanation:
From the picture attached,
lines 'l' ad 'm' are the parallel lines and line 't' is a transversal intersecting each line at two different points.
From the angles formed at the points of intersection,
∠5 ≅ ∠8 [Vertical angles]
∠5 ≅ ∠4 [Alternate interior angles]
∠5 ≅ ∠1 [Corresponding angles]
Therefore, Options A, D and F will be the correct options.
Answer:
x=29
Step-by-step explanation:
(x+3)+(x-1)+(x+1)=90
x+3+x-1+x+1=90
3x+3=90
3x=90-3
x=87/3
x=29
x+3=32
x+1=30
x-1=28
Checking
(x+3)+(x-1)+(x+1)=90
(29+3)+(29-1)+(29+1)=90
32+28+30=90
90=90