Remember, when you distribute exponents, you distribute them to the existing exponent (if there is one) AND the coefficient. Here's the math:
1. [tex](5^3)^5[/tex]
Distribute [tex](^5)[/tex] to 5 and [tex](^3)[/tex]
[tex](5^5)^3^(^5^)[/tex]
Multiply the terms accordingly
[tex]25^1^5[/tex]
2. [tex]\frac{7^-^2}{5^-^2}[/tex]
Multiply the numerator and denominator by
[tex]\frac{7^-^2}{5^-^2}=( \frac{7}{5})^-^2[/tex]
Use the Negative Exponent Rule to rearrange the expression
[tex](\frac{5}{7})^2[/tex]
Square the numerator and denominator
[tex]\frac{5^2}{7^2}[/tex]
Simplify
[tex]\frac{25}{49}[/tex]
3. [tex]-3(5+27)^0 = 0[/tex]
If something is multiplied to the power of 0, than it is equal to 0
4. [tex]10^-^2[/tex]
For this, we will once again use the Negative Exponents Rule (find the reciprocal and change the negative exponent to its opposite)
[tex] \frac{1}{10^2} [/tex]
Simplify
[tex]\frac{1}{100}[/tex]
Answers:
1. [tex]25^1^5[/tex]
2. [tex]\frac{25}{49}[/tex]
3. [tex]0[/tex]
4. [tex]\frac{1}{100}[/tex]
Hope this helps!
