This is simple. If you add 5 and 8 together, you get 13. What you do, is add the 1 to the 2, you get 3. The 3 in the 13 stays the same, so the awnser is 33.
Hello :
let A(0,3,2) and (Δ) this line , v vector parallel to (<span>Δ).
M</span>∈ (Δ) : vector (AM) = t v..... t ∈ R
1 ) (Δ) parallel to the plane x + y + z = 5 : let : n an vector <span>perpendicular
to the plane : n </span>⊥ v .... n(1,1,1) so : n.v =0 means : n.vector (AM) = 0
(1)(x)+(1)(y-3)+(1)(z -2) =0 ( vector (AM) = ( x, y -3 , z-2 )
x+y+z - 5=0 ...(1)
2) (Δ) perpendicular to the line (Δ') : x = 1+t , y = 3 - t , z = 2t :
vector (u) ⊥ v .... vector(u) parallel to (Δ') and vector(u) = (1 , -1 ,1)
vector (u) ⊥ vector (AM) means :
(1)(x)+(-1)(y-3)+(2)(z -2) =0
x - y+2z - 1 = 0 ...(2)
so the system :
x+y+z - 5=0 ...(1)
x - y+2z - 1 = 0 ...(2)
(1)+(2) : 2x+3z - 6 =0
x = 3 - (3/2)z
subsct in (1) : 3 - (3/2)z +y +z - 5 =0
y = 1/2z +2
let : z=t
an parametric equations for the line (Δ) is : x = 3 - (3/2)t
y = (1/2)t +2
z=t
verifiy :
1) (Δ) parallel to the plane x + y + z = 5 :
(-3/2 , 1/2 ,1) <span>perpendicular to (1,1,1)
</span>because : (1)(-3/2)+(1)(1/2)+(1)(1) = -1 +1 = 0
2) (Δ) perpendicular to the line (Δ') :
(-3/2 , 1/2 ,1) perpendicular to (1,-1,2)
because : (1)(-3/2)+(-1)(1/2)+(1)(2) = -2 +2 = 0
A(0, 3, 2)∈(Δ) :
0 = 3-(3/2)t
3 = (1/2)t+2
2 =t
same : t = 2
The base of the triangle is 12 because area of a triangle is 1/2 base times height
7-7
___
8-(-3)
0
_
11
Slope =0
The solution to the compound inequality given as 6b < 36 or 2b + 12 > 6 is b < 6 or b > -3
<h3>How to solve the
compound inequality?</h3>
The compound inequality is given as:
6b < 36 or 2b + 12 > 6
Evaluate the like terms in the individual inequalities
6b < 36 or 2b > -6
Divide the individual inequalities by the coefficients of b
b < 6 or b > -3
Hence, the solution to the compound inequality given as 6b < 36 or 2b + 12 > 6 is b < 6 or b > -3
Read more about compound inequality at
brainly.com/question/1485854
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