Answer:
4a-b-c
Step-by-step explanation:
we first look at the letters of each number. if there is only a letter, there is an imaginary 1 in front of it. we need to make sure that we add the numbers that have the exact same letters because they are known as like terms. we cannot add 6a and 4b because the letters are different and they are known as unlike terms. When we add numbers with the same letter ONLY the number changes
Answer:
$27.50
Step-by-step explanation:
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
It is 5 because it is an odd number and 5 factors are 5
