If U is midpoint of KE and LT , prove ∆KLU ≅ ΔETU. Statement Reason 1. KU ≅ 1. Def. of midpoint 2. LU ≅ 2. Def. of midpoint 3. ∠KUL ≅ ∠ 3. 4. △KLU ≅ △ 4.

тgє qυєѕтισи αѕкє∂: ιи тнє ѕєяιєѕ σя яєα¢тισиѕ тgαт σ¢¢υя ιи ρнσтσѕуитнєѕιѕ, ωнι¢н fσямѕ ωнєи ℓιgнт ѕтяιкєѕ α мσℓє¢υℓє σf ¢нℓσяσρнуℓℓ.

α. мσℓє¢υℓє σf ωαтєя

в.тнєє-¢αявσи ѕυgαя

¢. "єχ¢ιтє∂" єℓє¢тяσи

∂. иα∂ρн

му αиѕωєя: ιт ιѕ ¢. "єχ¢ιтє∂" єℓє¢тяσи

нσρє ι ¢συℓ∂ нєℓρ υ

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

its c