macar0ni

28/4 = 7, which is one of the rhombus's sides. since we know PR is 8, PT would be 4, so now we can use Pythagorean Theorem to solve for QT.

7^2 = b^2+4^2

49 = b^2+16

33 = b^2

b = \(\sqrt33\)

i hope this helped! please correct me if i am wrong:)

nice, CPhill! but when you solved for f(f(x)), did you maybe forget the 1/ part of the function?

and did you maybe do 2x+5 instead of 2x-5? or maybe you just did some simplifying and i didn't follow that, sorry if that is what you did and i just didn't notice ;o

expanding:

(2x+1/6)(2x+1/6)

4x^2+x/3+x/3+1/36

combining like terms:

3x^2+2x/3+1/36

i hope this helped! please correct me if i am wrong:)

the sum of the interior angles of a triangle is 180 degrees, so:

3x+1+2x-3+9x = 180

simplifying,,

14x-2 = 180

14x = 182

x = 13

the obvious smallest angle is the 2x-3 one, so plugging our x in:

2(13) - 3 = 26 - 3 = 23

so, the answer is 23 degrees. 

hope this helped! please correct me if i am wrong:)

whoops, thank you for catching that :O

my bad, y should be 153/6, sorry about that !!

(...and i hope you can solve for x on your own:))

we can set up two equations here:

x = 5y - 13

x + y = 140

now, we can use substitution:

5y - 13 + y = 140

6y - 13 = 140

6y = 173

y = 173/6

x = 667/6

hope this helped! please correct me if i am wrong:)

suubtracting 2x from both sides, you get:

x = 12

hope this helped! :)

i get it now, thank you so much!

yes, Vockey, i think that is right! :)

very nice, vockey!