$e\cdot e^x -e^{-2}=-2$
$\implies e^{x+1}=e^{-2}-2$
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
Answer:
p = -12
Step-by-step explanation:
I used graph paper to graph point (-7,-6) then i used the slope to continue to graph the rest of the points until i got to (-1,-12).
Answer:
see attached
Step-by-step explanation:
The equation is in the form ...
4p(y -k) = (x -h)^2 . . . . . (h, k) is the vertex; p is the focus-vertex distance
Comparing this to your equation, we see ...
p = 4, (h, k) = (3, 4)
p > 0, so the parabola opens upward. The vertex is on the axis of symmetry. That axis has the equation x=x-coordinate of vertex. This tells you ...
vertex: (3, 4)
axis of symmetry: x = 3
focus: (3, 8) . . . . . 4 units up from vertex
directrix: y = 0 . . . horizontal line 4 units down from vertex
Answer:
4 1/3
Step-by-step explanation:
Hope this helps :)