How many $c m^{2}$ $cm^2$ are there in $4.21 \cdot {inch}^{2}$ $4.21*"inch"^2$ ?

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How many $c m^{2}$ $cm^2$ are there in $4.21 \cdot {inch}^{2}$ $4.21*"inch"^2$ ?

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Answer 1 · 2017-08-30T02:47:24

Well, $1 \cdot inch \equiv 2.54 \cdot c m$ $1*"inch"-=2.54*cm$ ..........

Explanation:

And so, $4.21 \cdot {inch}^{2} = 4.21 \cdot {inch}^{2} \times {2.54 \cdot c m \cdot {inch}^{- 1}}^{2}$ $4.21*"inch"^2=4.21*"inch"^2xx{2.54*cm*"inch"^-1}^2$

$= 27.2 \cdot c m^{2}$ $=27.2*cm^2$

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How many $c m^{2}$ $cm^2$ are there in $4.21 \cdot {inch}^{2}$ $4.21*"inch"^2$ ?

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