Jawaban:
[tex]\begin{gathered}\boxed{\begin{array}{c|c}\bf Bentuk \: akar & \bf Hubungan \: bilangan \: Berpangkat \: \\& \bf dengan \: Bentuk \: Akar\\\hline \sqrt{25} & \sf 25 = {5}^{2} \to \sqrt{25} = 5 \\\hline \sqrt{36} & \sf 36 = \red{ {6}^{2}} \to \sqrt{36} = \red{6} \\\hline \sqrt{81} & \sf 81 = \red{ {9}^{2}} \to \sqrt{81} = \red{9}\\\hline \sqrt{100} & \sf 100 = \red{ {10}^{2}} \to \sqrt{100} = \red{10} \\\hline \sqrt{144} & \sf 144 = \red{ {12}^{2}} \to \sqrt{144} = \red{12} \\\hline \sqrt{225} & \sf 225 = \red{ {15}^{2}} \to \sqrt{225} = \red{15}\end{array}}\end{gathered}[/tex]
'비상' (Svt)
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Jawaban:
Penyelesaian :
[tex]\begin{gathered}\boxed{\begin{array}{c|c}\bf Bentuk \: akar & \bf Hubungan \: bilangan \: Berpangkat \: \\& \bf dengan \: Bentuk \: Akar\\\hline \sqrt{25} & \sf 25 = {5}^{2} \to \sqrt{25} = 5 \\\hline \sqrt{36} & \sf 36 = \red{ {6}^{2}} \to \sqrt{36} = \red{6} \\\hline \sqrt{81} & \sf 81 = \red{ {9}^{2}} \to \sqrt{81} = \red{9}\\\hline \sqrt{100} & \sf 100 = \red{ {10}^{2}} \to \sqrt{100} = \red{10} \\\hline \sqrt{144} & \sf 144 = \red{ {12}^{2}} \to \sqrt{144} = \red{12} \\\hline \sqrt{225} & \sf 225 = \red{ {15}^{2}} \to \sqrt{225} = \red{15}\end{array}}\end{gathered}[/tex]
'비상' (Svt)
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