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\[ \left(3 k^{5} n^{2}\right)\left(-4 k^{6} n^{3}\right) \] implify the following expressions a there possible: \( \left(-2 k^{2} m\right)\left(-8 k^{9} m^{2}\right) \) \( 6 a^{3} b^{2} \times 2 a^{4} b^{8} \times a^{2} \) \( x^{6} y^{3} z^{2} \cdot x^{2} z^{4} y^{3}\left(2 x^{5} \hbar^{4}\right) \)

Ask by Curry Mckenzie. in Eswatini
Mar 17,2025

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\[ -12k^{11}n^{5}, \quad 16k^{11}m^{3}, \quad 12a^{9}b^{10}, \quad 2x^{13}y^{6}z^{6}\hbar^{4} \]

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The Deep Dive

To simplify the expression \((3 k^{5} n^{2})(-4 k^{6} n^{3})\), you just multiply the coefficients and add the exponents of like bases. The coefficient becomes \(3 \times -4 = -12\), and for \(k\), \(k^{5} \cdot k^{6} = k^{11}\). For \(n\), \(n^{2} \cdot n^{3} = n^{5}\). So the simplified result is \(-12 k^{11} n^{5}\). Now for the second expression \((-2 k^{2} m)(-8 k^{9} m^{2})\), we again multiply the coefficients to get \(16\) (since \(-2 \times -8 = 16\)). For \(k\), we have \(k^{2} \cdot k^{9} = k^{11}\), and for \(m\), \(m^{1} \cdot m^{2} = m^{3}\). Thus, the final result is \(16 k^{11} m^{3}\). Next, for \(6 a^{3} b^{2} \times 2 a^{4} b^{8} \times a^{2}\), multiply the coefficients to get \(6 \times 2 = 12\), and for \(a\), combine to get \(a^{3} \cdot a^{4} \cdot a^{2} = a^{9}\). For \(b\), we have \(b^{2} \cdot b^{8} = b^{10}\). Therefore, the simplified form is \(12 a^{9} b^{10}\). Finally, for \(x^{6} y^{3} z^{2} \cdot x^{2} z^{4} y^{3}(2 x^{5} \hbar^{4})\), we multiply the coefficients to get \(2\), similar to adding exponents for \(x\), \(y\), and \(z\): \(x^{6} \cdot x^{2} \cdot x^{5} = x^{13}\), \(y^{3} \cdot y^{3} = y^{6}\), and \(z^{2} \cdot z^{4} = z^{6}\). Thus, the simplified result is \(2 x^{13} y^{6} z^{6} \hbar^{4}\).

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