Answer:
tbh i like the first one its so simple but like so nice at the same time
?
Step-by-step explanation:
2 ways: Easy and hard
Hard=A
Easy=B
A: 1/2x+4
work from there so we do fun stuff with it
make something that can be simplified so
1/2x+4 times (2/2)=x+8
now square the whole thing and put the result in a square root thingie
(x+8)^2=x^2+16x+64

multiply the whole thing by 4/4 and put
![\sqrt{16} [\tex] on top so then [tex] \sqrt{x^2+16x+64}](https://tex.z-dn.net/?f=%20%5Csqrt%7B16%7D%20%5B%5Ctex%5D%20on%20top%20so%20then%20%0A%5Btex%5D%20%5Csqrt%7Bx%5E2%2B16x%2B64%7D%20)
times

=

=

to solve it, factor out the 16 in the square root and then square root 16 to get 4
then it will be (4 times square root of equation)/4=square root of equatio
factor square root of equation and square root it and get x+8
divide by 2 to get 1/2x+4
B: 1/2x+4
put stuff that cancels out
1/2x+3x-3x+4+56-56
move them around
3 and 1/2x-3x+60-56
or
2x-3x+1 and 1/2x+30-20+30-36
then just add like terms to solve
Answer:
Step-by-step explanation:
Given is a triangle RST and another triangle R'S'T' tranformed from RST
Vertices of RST are (0, 0), (negative 2, 3), (negative 3, 1).
Vertices of R'S'T' are (2, 0), (0, negative 3), (negative 1, negative 1).
Comparing the corresponding vertices we find that x coordinate increased by 2 while y coordinate got the different sign.
This indicates that there is both reflection and transformation horizontally to the right by 2 units
So first shifted right by 2 units so that vertices became
(2,0) (0,3) (-1,1)
Now reflected on the line y=0 i.e. x axis
New vertices are
(2,0) (0,-3) (-1,-1)
The area of a triangle is A = (b*h)/2:
(5 * 8)/2 = 20