A alternativa correta é; 1) C).
A área do quadrado que Sebastião desenhou foi, 2) D).
[tex]\boxed{\begin{array}{lr} \dfrac{5}{3}x^6y^4 \end{array}}\\\\\\\boxed{\begin{array}{lr} \dfrac{5}{3}\ .\ \dfrac{5}{3} =\dfrac{5\ .\ 5}{3\ . 3}=\boxed{\begin{array}{lr} \dfrac{25}{9} \end{array}}\end{array}}\\\\\\\boxed{\begin{array}{lr} x^6\ .\ x^6=\boxed{\begin{array}{lr} x^1^2 \end{array}}\end{array}}\\\\\\\boxed{\begin{array}{lr} y^4 \ . \ y^4=y^8 \end{array}}[/tex]
Resposta;
Logo, será a alternativa C).
[tex]\boxed{\begin{array}{lr} \boxed{\begin{array}{lr} \boxed{\begin{array}{lr} \boxed{\begin{array}{lr} \boxed{\begin{array}{lr} \dfrac{25}{9}\ x^{12}\ y^{8}\ \ \ \checkmark \end{array}} \end{array}} \end{array}} \end{array}} \end{array}}[/tex]
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A = L . L
A = L²
A = (10x)²
A = 100x²
Resposta;
D) 100x²
Respostas;
C) e D).
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