Give an example of a monotone simple polygon for which the set of directions with respect to which
it is monotone consists of at least two distinct "double cones" of directions. more precisely, show a simple polygon p that
is monotone with respect to the line y = x and with respect to the line y = −x, but is not x-monotone or y-monotone. 1
(optional question to provoke thought: for an n-gon, what is the maximum number (in terms of n) of double cones of
directions of monotonicity? i know that it can grow like ω(n), but can you get a tight exact bound? )
(b). suppose that p is a simple rectilinear polygon (all edges are parallel to the coordinate axes). ca

(d)

step-by-step explanation:

given: the line in the graph passes through points (5,30) and (6,26).

then, the equation of line is:

now, put y=a(n) and x=n,

hence, option d is correct.

the final price would be 29.6925

step-by-step explanation:

39.59 minus 25% equals 29.6925

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!