ROOTS AND CUBES
Can someone please help me answer all 4 following?




NOTES:
Answer: (C) 41
Step-by-step explanation:
[tex]\quad 6^2+\sqrt[3]{125} \\= 6 \cdot 6+\sqrt[3]{5\cdot 5 \cdot 5}\\= 36 + 5\\= 41[/tex]
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Answer: (C) 10
Step-by-step explanation:
[tex]\bigg(\dfrac{7}{3}\times \sqrt[3]{27}-2\bigg)\times \dfrac{1}{5} + \sqrt{81}[/tex]
[tex]=\bigg(\dfrac{7}{3}\times \sqrt[3]{3\cdot 3 \cdot 3}-2\bigg)\times \dfrac{1}{5} + \sqrt{9\cdot 9}[/tex]
[tex]=\bigg(\dfrac{7}{3}\times3-2\bigg)\times \dfrac{1}{5}+9[/tex]
[tex]=(7 - 2)\times \dfrac{1}{5}+9[/tex]
[tex]=5 \times \dfrac{1}{5}+ 9[/tex]
[tex]= 1 + 9[/tex]
[tex]= 10[/tex]
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8² = 8 · 8 = 64 11² = 11 · 11 = 121
5³ = 5 · 5 · 5 = 125 3³ = 3 · 3 · 3 = 27
[tex]\sqrt{1600}=\sqrt{40\cdot 40}=40[/tex] [tex]\sqrt[3]{64}=\sqrt[3]{4\cdot 4\cdot 4}=4[/tex]
[tex]\sqrt{144}=\sqrt{12\cdot 12}=12[/tex] [tex]\sqrt[3]{8000}=\sqrt[3]{20\cdot 20\cdot 20}=20[/tex]