) Prove that the following function f is injective (one-to-one): f:Z→Z, where f(x)=3x+2 Do this using the contrapositive of the definition of injective: Show that "for all a, b in the domain (ℤ), if f(a) = f(b) then a = b."

answer: 5 17/33 % (to the tenth)

step-by-step explanation:

he guessed: 34.5 pounds

actual weight is 32.7

there for he is off:

34.5 -32.7 = 1.8

32.7 is the 100%

1.8 represents %? of 32.7

therefore:

32.7   - 100%

1.8     -     x%

x= 1.8*100%/32.7% = 18*100%/327%

x=1800%/327%   = (5 165/327)%

to the tenth:

5 17/33 %

asnwer is (g o h o f)(x) = 2x - 6/x + 7

step-by-step explanation:

given,

f(x) = x - 3/x

g(x) = x + 3

h(x) = 2x + 1

we need to find : (g o h o f)(x)

we can write (g o h o f)(x) = g[ h{ f(x) } ]

so first we find h{f(x)}

h{f(x)} = f(x) + 3 = (x - 3/x) + 3

h{f(x)} = x - 3/x + 3

now we find g[ h{ f(x) } ].

g[ h{ f(x) } ] = g(x - 3/x + 3)       {plug h{f(x)} = x - 3/x + 3 in g[h{f(x)}] }

g(x - 3/x + 3) = 2 (x - 3/x + 3) + 1

                    = 2x - 6/x + 6 + 1

                    = 2x - 6/x + 7

hence we get (g o h o f)(x) = 2x - 6/x + 7

i believe so, according to my "strange" way of figuring out weird math problems like this.