Question
Making use of the cube root table, find the cube root $7800$
✓
Answer
We have: $7800 = 78 \times 100 \therefore\sqrt[3]{7800}$
$=\sqrt[3]{78\times100}$
$=\sqrt[3]{78}\times\sqrt[3]{100}$ By the cube root table,
we have: $=\sqrt[3]{78}=4.273$ and $\sqrt[3]{100}=4.642$
$\therefore\sqrt[3]{7800}$
$=\sqrt[3]{78}\times\sqrt[3]{100}$
$=4.273\times4.642$
$=19.835$ (Up to three decimal places)
Thus, the answer is $19.835.$
$=\sqrt[3]{78\times100}$
$=\sqrt[3]{78}\times\sqrt[3]{100}$ By the cube root table,
we have: $=\sqrt[3]{78}=4.273$ and $\sqrt[3]{100}=4.642$
$\therefore\sqrt[3]{7800}$
$=\sqrt[3]{78}\times\sqrt[3]{100}$
$=4.273\times4.642$
$=19.835$ (Up to three decimal places)
Thus, the answer is $19.835.$
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