So what we do is
area that remains=total area-triangle area that was cut out
we need to find 2 things
total area
triangle area
total area=rectange=base times height
area=(3x+4) times (2x+3)
FOIL or distribute
6x^2+8x+9x+12=6x^2+17x+12
triangle area=1/2 times base times height
triangle area=1/2 times (2x+2) times (x-2)=
(x+2) times (x-2)=x^2+2x-2x-4=x^2-4
so
total area=6x^2+17x+12
triangle area=x^2-4
subtract
area that remains=total area-triangle area that was cut out
area that remains=6x^2+17x+12-(x^2-4)=
6x^2+17x+12-x^2+4=
6x^2-x^2+17x+12+4=
5x^2+17x+16
area that remains is 5x^2+17x+16
I know that B is orthographic.
D is isometric.
A is a net.
C is none.
Answer:
Step-by-step explanation:
Answer:
A) JK and AB =
![[tex] \boxed{( \theta) = \f{12}{( \theta)} }](https://tex.z-dn.net/?f=%5Btex%5D%20%5Cboxed%7B%28%20%5Ctheta%29%20%3D%20%5Cf%7B12%7D%7B%28%20%5Ctheta%29%7D%20%7D)
So, =》 
=》 
[/tex]
B) AABC
![[tex] \boxed{( \theta) = {1}{( \theta)} }](https://tex.z-dn.net/?f=%5Btex%5D%20%5Cboxed%7B%28%20%5Ctheta%29%20%3D%20%7B1%7D%7B%28%20%5Ctheta%29%7D%20%7D)
So, 
C) AABM:
![[tex] e \fbox{: }](https://tex.z-dn.net/?f=%5Btex%5D%20e%20%5Cfbox%7B%3A%20%7D)
The value of k for which 
is 
ans[/tex]
----------
Answer:
f equals 85
Step-by-step explanation:
85 times 5 equals 425 plus 25 equal to 450
