Tienes tu uno negro Thousand cris
16 is the answer (I used a triangle calculator)
Let's buils the intersection plane:
Point P is on AB and AP=2, then PB=3; point Q is on AE and AQ=1, then QE=4. Let P' be a point on CD such that CP'=2 and Q' be a point on the plane CDHG such that P'Q'=1 and P'Q' is perpendicular to CD. The line CQ' intersects HD at point R and the plane CPQR is intersection plane.
Consider triangles ΔCDR and ΔCP'Q', they are similar. So,
,
so R is a midlepoint of the side HD (for details see picture).
Answer:
Step-by-step explanation:
a) The graph rises steeply from the third quadrant and there is a point of inflection through the origin (0, 0) then rises steeply in quadrant 1.
b) Let y = g(x) = x^3
x = ∛y
So g(-1)x = ∛x.
The graph is the reflection of g(x) in y = x.
c) Domain of g(x) = All real x, Range is all real g(x).
Domain of g(-1)x = All real x, Range = all real g-1(x).
d) You will see the end behaviour from the graphs.
I tried to send you a link to the graphs but the system wont allow it.