"given: in ∆abc, segment de is parallel to segment ac .. prove: bd over ba equals be over bc. the two-column proof with missing statements and reasons proves that if a line parallel to one side of a triangle also intersects the other two sides, the line divides the sides proportionally.. statement. reason. 1. segment de is parallel to segment ac. 1. given. 2. line segment ba is a transversal that intersects two parallel lines. 2. conclusion from statement 1. 3.. 3.. 4. ∡bde ≅ ∡bac. 4. reflexive property of equality. 5. ∆abc ~ ∆dbe. 5. angle-angle (aa) similarity postulate. 6.. 6.. "

47

step-by-step explanation:

after simplifying it i got: -36.25 as my answer. hope that !

l.ln.lln.k.nkl

step-by-step explanation:

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!