Ella completed the following work to test the equivalence of two expressions.

Question

5f+3.6
5(1)+3.6
5+3.6
8.6

6f+2.6
6(1)+2.6
6+2.6
8.6

Which is true about the expressions?
(1 point)

A. The expressions are equivalent because Ella got different results when she substituted one for f.

B. The expressions are not equivalent because Ella would get the same results when substituting different numbers for f.

C. The expressions are equivalent because Ella got the same result when she substituted one for f.

D. The expressions are not equivalent because Ella would get different results when substituting different numbers for f.

Answer 1

The correct option is:

C. The expressions are equivalent because Ella got the same result when she substituted one for f.

Ella substituted \( f = 1 \) in both expressions and obtained the same result of \( 8.6 \), which indicates that the expressions are equivalent.

Answer 2

you are wrong

Answer 3

I apologize for any confusion. Let’s clarify the work done by Ella:

1. For the first expression \( 5f + 3.6 \):

   • When Ella substituted \( f = 1 \):
     • \( 5(1) + 3.6 = 5 + 3.6 = 8.6 \)

2. For the second expression \( 6f + 2.6 \):

   • When Ella substituted \( f = 1 \):
     • \( 6(1) + 2.6 = 6 + 2.6 = 8.6 \)

Since both expressions yield the same result of \( 8.6 \) when \( f = 1 \), we might consider them equivalent at that point. However, to fully determine if the expressions are equivalent for all values of \( f \), we can express the two equations generally:

• \( 5f + 3.6 \)
• \( 6f + 2.6 \)

If we compare the coefficients of \( f \) and the constant terms, we see:

1. \( 5f + 3.6 \) has a coefficient of 5 for \( f \) and a constant of \( 3.6 \).
2. \( 6f + 2.6 \) has a coefficient of 6 for \( f \) and a constant of \( 2.6 \).

Because the coefficients of \( f \) in each expression differ, these expressions are not equivalent in general (outside of the specific value \( f = 1 \)).

Therefore, the correct answer is:

D. The expressions are not equivalent because Ella would get different results when substituting different numbers for \( f \).

Thank you for your patience, and I'm glad to clarify!