Answer/Step-by-step explanation:
✔️<1 can be named in two ways using letters, with the middle letter being the letter of the vertex of the angle. Thus is can be named as:
or as 
✔️It is unclear to name <1 as <D because <2 can equally be named as <2 since both angles share a common point or vertex D.
The coordinate pair can be obtained from the given equation by
rearranging the equation to make <em>y</em> the subject of the equation
- The table of values obtained from the given equation is presented as follows;
- The graph showing the points in the above table is attached
Reason:
The given equation is presented as follows;
2·x - 4·y = 12
(a) The values are found by making either <em>x</em>, or <em>y</em> the subject of the
formula, and inputting values as follows;
2·x - 4·y = 12
= 4·y
4·y = 2·x - 12
The rearranged equation is therefore
Plugging in values of <em>x</em>, (0, 1, 2, 3) to find the corresponding y-values gives;
x = 0
x = 3
x = 4
<u>The table of values</u> is therefore;
![\begin{array}{|c|cr|} \mathbf{x}&&\mathbf{y}\\0&&-3\\1&&-2.5\\2&&-2\\3&&-1.5\\4&&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cc%7Ccr%7C%7D%20%5Cmathbf%7Bx%7D%26%26%5Cmathbf%7By%7D%5C%5C0%26%26-3%5C%5C1%26%26-2.5%5C%5C2%26%26-2%5C%5C3%26%26-1.5%5C%5C4%26%26-1%5Cend%7Barray%7D%5Cright%5D)
The <u>graph of the points</u> can be <u>plotted</u> using MS Excel as presented here
Learn more here:
brainly.com/question/7221077
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
