$e\cdot e^x -e^{-2}=-2$
$\implies e^{x+1}=e^{-2}-2$
note that RHS is negative. (because e with negative exponent is less than 1)
and LHS is always positive.
so there cannot be any solution
We have:
The generic equation of the line is: y-yo = m (x-xo)
The slope is:
m = (y2-y1) / (x2-x1)
m = (- 2-0) / (3-0)
m = -2 / 3
We choose an ordered pair
(xo, yo) = (0, 0)
Substituting values:
y-0 = (- 2/3) (x-0)
Rewriting:
y = (- 2/3) x
Answer:
The equation of the line is:
y = (- 2/3) x
Answer:
c^2=33^2+37^2-2(33)(37)cos120(i don't know for sure tho)
Step-by-step explanation:
LAW OF COSINES
Step-by-step explanation:
Answer attached :)
Hope it helps :D
If h moves the graph left or right,
(moves left)
(moves right)
If a vertical stretch by a factor of |h|, then
If h moves the graph up or down,
(moves up)
(moves down)
and h = 1