The figure below shows a rectangle abcd having diagonals ac and db: a rectangle abcd is shown with diagonals ac and bd. jimmy wrote the following proof to show that the diagonals of rectangle abcd are congruent: jimmy's proof: statement 1: in triangle adc and bcd, ad = bc (opposite sides of a rectangle are congruent) statement 2: angle adc = angle bcd (angles of a rectangle are 90°) statement 3: statement 4: triangle adc and bcd are congruent (by sas postulate) statement 5: ac = bd (by cpctc) which statement below completes jimmy's proof? ab=ab (reflexive property of equality) ab=ab (transitive property of equality) dc=dc (reflexive property of equality) dc=dc (transitive property of equality)

C. DC = DC. By reflexive property of equality.

Step-by-step explanation:

We are given that,

In step 4, Jimmy proved ΔADC ≅ ΔBCD by the SAS Congruence Postulate.

For that, the previous steps states,

Step 1: AD = BC, as opposite sides of a rectangle are congruent.

Step 2: ∠ADC = ∠BCD

So, in order to satisfy the SAS Postulate, we must have another pair of sides equal from the respective rectangles.

As, we can see that in ΔADC and ΔBCD, the side DC is a common side.

Thus, in the missing step, we get,

Step 3: DC = DC, By the reflexive property of equality.

Hence, option C is correct.

C. DC = DC. By reflexive property of equality.

Step-by-step explanation:

Just took the quiz, and got this as the correct answer.

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