Which of these is equal to [tex]$5^0 + 2^1$[/tex]?

A. [tex]$2^0 + 1^1$[/tex]
B. [tex][tex]$2^0 + 2^1$[/tex][/tex]
C. [tex]$6^0 - 3^0$[/tex]
D. [tex]$6^0 - 6^1$[/tex]



Answer :

Let's evaluate each of the expressions step-by-step:

1. Evaluate [tex]\(5^0 + 2^1\)[/tex]:
[tex]\[ 5^0 = 1 \quad \text{(Any non-zero number to the power of zero is 1)} \][/tex]
[tex]\[ 2^1 = 2 \quad \text{(Any number to the power of one is itself)} \][/tex]
Hence:
[tex]\[ 5^0 + 2^1 = 1 + 2 = 3 \][/tex]

2. Evaluate [tex]\(2^0 + 1^1\)[/tex]:
[tex]\[ 2^0 = 1 \quad \text{(Any non-zero number to the power of zero is 1)} \][/tex]
[tex]\[ 1^1 = 1 \quad \text{(Any number to the power of one is itself)} \][/tex]
Hence:
[tex]\[ 2^0 + 1^1 = 1 + 1 = 2 \][/tex]

3. Evaluate [tex]\(2^0 + 2^1\)[/tex]:
[tex]\[ 2^0 = 1 \quad \text{(Any non-zero number to the power of zero is 1)} \][/tex]
[tex]\[ 2^1 = 2 \quad \text{(Any number to the power of one is itself)} \][/tex]
Hence:
[tex]\[ 2^0 + 2^1 = 1 + 2 = 3 \][/tex]

4. Evaluate [tex]\(6^0 - 3^0\)[/tex]:
[tex]\[ 6^0 = 1 \quad \text{(Any non-zero number to the power of zero is 1)} \][/tex]
[tex]\[ 3^0 = 1 \quad \text{(Any non-zero number to the power of zero is 1)} \][/tex]
Hence:
[tex]\[ 6^0 - 3^0 = 1 - 1 = 0 \][/tex]

5. Evaluate [tex]\(6^0 - 6^1\)[/tex]:
[tex]\[ 6^0 = 1 \quad \text{(Any non-zero number to the power of zero is 1)} \][/tex]
[tex]\[ 6^1 = 6 \quad \text{(Any number to the power of one is itself)} \][/tex]
Hence:
[tex]\[ 6^0 - 6^1 = 1 - 6 = -5 \][/tex]

Therefore, comparing all the values:

- [tex]\(5^0 + 2^1 = 3\)[/tex]
- [tex]\(2^0 + 1^1 = 2\)[/tex]
- [tex]\(2^0 + 2^1 = 3\)[/tex]
- [tex]\(6^0 - 3^0 = 0\)[/tex]
- [tex]\(6^0 - 6^1 = -5\)[/tex]

The expression that is equal to [tex]\(5^0 + 2^1\)[/tex] is [tex]\(2^0 + 2^1\)[/tex].