Asked by I_EAT_EVERYDAY
Which of these is correct?(1 point)
Responses:
4^0 over 4^3 = 0
6^7 over 6^−5 = 36
3^6 over 3^4 = 1 over 9
5^8 over 5^6 = 25
Responses:
4^0 over 4^3 = 0
6^7 over 6^−5 = 36
3^6 over 3^4 = 1 over 9
5^8 over 5^6 = 25
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To check the correctness of each statement, we can simplify each expression:
1. **\( \frac{4^0}{4^3} \)**:
\( 4^0 = 1 \), so \( \frac{4^0}{4^3} = \frac{1}{4^3} = \frac{1}{64} \), which is NOT equal to 0.
2. **\( \frac{6^7}{6^{-5}} \)**:
Using the property of exponents \( \frac{a^m}{a^n} = a^{m-n} \), we have:
\( \frac{6^7}{6^{-5}} = 6^{7 - (-5)} = 6^{7 + 5} = 6^{12} \). This is NOT equal to 36.
3. **\( \frac{3^6}{3^4} \)**:
Again using the property of exponents:
\( \frac{3^6}{3^4} = 3^{6 - 4} = 3^2 = 9 \). This is NOT equal to \( \frac{1}{9} \).
4. **\( \frac{5^8}{5^6} \)**:
Using the property of exponents:
\( \frac{5^8}{5^6} = 5^{8 - 6} = 5^2 = 25 \). This statement is correct.
So, the only correct statement is:
**\( 5^8 \) over \( 5^6 = 25 \)**.
1. **\( \frac{4^0}{4^3} \)**:
\( 4^0 = 1 \), so \( \frac{4^0}{4^3} = \frac{1}{4^3} = \frac{1}{64} \), which is NOT equal to 0.
2. **\( \frac{6^7}{6^{-5}} \)**:
Using the property of exponents \( \frac{a^m}{a^n} = a^{m-n} \), we have:
\( \frac{6^7}{6^{-5}} = 6^{7 - (-5)} = 6^{7 + 5} = 6^{12} \). This is NOT equal to 36.
3. **\( \frac{3^6}{3^4} \)**:
Again using the property of exponents:
\( \frac{3^6}{3^4} = 3^{6 - 4} = 3^2 = 9 \). This is NOT equal to \( \frac{1}{9} \).
4. **\( \frac{5^8}{5^6} \)**:
Using the property of exponents:
\( \frac{5^8}{5^6} = 5^{8 - 6} = 5^2 = 25 \). This statement is correct.
So, the only correct statement is:
**\( 5^8 \) over \( 5^6 = 25 \)**.
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