Segments bc and de are parallel. find x
a. 5
b. 4
c. 6
d. 7

C. 6  

Step-by-step explanation:

∆ABC and ∆ADE are similar triangles.

                       AB/AC = AD/AE

(x - 1 + x + 2)/(15 + 24) = (x - 1)/15

                 (2x + 1)/39 = (x - 1)/15         Multiply each side by 39

                        2x + 1 = 39(x - 1)/15     Multiply each side by 15

                   15(2x + 1) = 39(x - 1)         Distribute 15 and 39

                    30x + 15 = 39x - 39        Add 39 to each side

                   30x + 54 = 39x               Subtract 30x from each side

                              54 = 9x                Divide each side by 9

                                x = 6

Check:

(6 - 1 + 6 + 2)/(15 + 24) = (6 - 1)/15

                          13/39 = 5/15

                                ⅓ = ⅓

the answer is:

mean: 6

mad: 2.14

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