\( .10 \rightarrow 8 \) HW Score: \( 10.64 \%, 10 \) of 94 points \( \begin{array}{l}(7 y)^{9} \cdot(7 y)^{2}-7 y^{11} \\ (\text { Type exponential notation with positive exponents.) }\end{array} \)
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When you're simplifying the expression \( (7y)^{9} \cdot (7y)^{2} - 7y^{11} \), you can use the laws of exponents! First off, remember that when you multiply like bases, you add the exponents: \( (7y)^{9} \cdot (7y)^{2} = (7y)^{11} \). This means our expression becomes \( (7y)^{11} - 7y^{11} \). Now you've got a common factor in both terms! You can factor out \( 7y^{11} \), giving you \( 7y^{11}(1 - 1) \), which equals \( 7y^{11} \cdot 0 = 0\). So the whole expression simplifies down to just \( 0 \). Simple as pie! 🥧