How do you multiply $(3 k^{4} d^{6}) (6 d^{3} j^{4})$ $(3k ^ { 4} d ^ { 6} ) ( 6d ^ { 3} j ^ { 4} )$ ?

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How do you multiply $(3 k^{4} d^{6}) (6 d^{3} j^{4})$ $(3k ^ { 4} d ^ { 6} ) ( 6d ^ { 3} j ^ { 4} )$ ?

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Answer 1 · 2017-02-24T13:05:41

Using indices first law, you can multiply the two d units together. $3 k^{4} \cdot 6 d^{6 + 3} \cdot j^{4}$ $3k^4*6d^(6+3)*j^4$ and thus the final answer would be $18 k^{4} d^{9} j^{4}$ $18k^4d^9j^4$

Explanation:

The trick is to put the d units together. $6 d^{3} \cdot d^{6}$ $6d^3*d^6$ would equal to $6 \cdot d^{3 + 6} = 6 d^{9}$ $6*d^(3+6)=6d^9$

Grouping using brackets where appropriate to highlight the relationships.

$(3 \times 6) (k^{4}) (d^{6} d^{3}) (j^{4})$ $(3xx6)(k^4)(d^6d^3)(j^4)$

$18 k^{4} d^{9} j^{4}$ $18k^4d^9j^4$

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How do you multiply $(3 k^{4} d^{6}) (6 d^{3} j^{4})$ $(3k ^ { 4} d ^ { 6} ) ( 6d ^ { 3} j ^ { 4} )$ ?

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How do you multiply $(3 k^{4} d^{6}) (6 d^{3} j^{4})$ $(3k ^ { 4} d ^ { 6} ) ( 6d ^ { 3} j ^ { 4} )$ ?