Which polynomial function has a root of 3 with multiplicity 2 and a root of 6 with multiplicity 1? f(x) = (x – 3)(x – 6)
f(x) = 2(x – 3)(x – 6)
f(x) = (x – 3)(x – 3)(x – 6)
f(x) = (x – 3)(x – 6)(x – 6)

ιт ιѕ в. 1/4

ιи (fяα¢тισи/єχα¢т fσям) 1/4

ιи (∂є¢ιмαℓ fσям) .25

нσρє ι ¢συℓ∂ нєℓρ уα

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

m∠abd = m∠dbc

step-by-step explanation:

given : ab bisects ∠dbc

as we know that bisection means ab divides   ∠dbc   in two equal angles named as   ∠dba and   ∠abc

so, m∠abd = m∠dbc  

because ab bisects ∠dbc.

hence , the first statement m∠abd = m∠dbc is true.

you didn't give an question you just gave us a statement