If the original square had a side length of
x = 2 inches, then what is the area of the
second rectangle? show how you arrived at
your answer.

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas  in the attached figure N 3, and the trinomial is

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas  in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

The total area of the second rectangle is the sum of the four areas

State the product of (x+4) and (x+7) as a trinomial

Part c) If the original square had a side length of  x = 2 inches, then what is the area of the  second rectangle?

we know that

The area of the second rectangle is equal to

For x=2 in

substitute the value of x in the area of each portion

Part d) Verify that the trinomial you found in Part b) has the same value as Part c) for x=2 in

We have that

The trinomial is

For x=2 in

substitute and solve for A(x)

----> verified

therefore

The trinomial represent the total area of the second rectangle

yes.. it's proven to be 36.

37 = ending with 7

unit's digit in the cube..3

56 = ending with 6

unit's digit in the cube..2

133 = ending with

unit's digit in the cube..7

hope it : ( : )