Hi there!
f(x) = 2x + 6
Differentiating 2x = 2 (Power rule, dy/dx xⁿ = nxⁿ⁻¹)
Differentiating a constant always equals 0.
Thus:
f'(x) = 2
Answer:
2.3 x = 10.4 - 2.9 x
2.3 x - 10.4 = 2.9 x
5.2 x -1.6 = 8.8
Step-by-step explanation:
Answer:
I believe it is B.
Step-by-step explanation:
y=mx+b
y=mx+3
y= -1/2x+3
Answer: 
Step-by-step explanation:
The quadratic equation in its standard form is:
Now, we are given 5 points of the parabola (if you graph a quadratic equation you will have a parabola), however we only need to choose three points to find the coeficients 
, 
and 
in the quadratic equation.
So, let's choose the first three points:
(-1,14):
(1)
(0,7):
(2)
(1,4):
(3)
Substituting (2) in (1) and (3):
(4)
(5)
At this point we have a system with two equations.
Adding (4) to (5):
(6)
Isolating :
(7)
Substituitng (7) in (3):
(8)
Isolating :
(9)
Now we have the three coeficients and we can write the quadratic equation:
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
