Question

solve the following given above​

Answer 1

Step-by-step explanation:

$(a) \: = 40 {m}^{2 - 2} \times {n}^{3 + 12} \times {p}^{ - 2 + 14 } \\ = 40 {n}^{15} {p}^{12} \\ (b) \: = ({4 {a}^{ - 5 - 1} . {b}^{3 + 12}. {c}^{ - 8} })^{ - 3} \\ = {(4 {a}^{ - 6} . {b}^{15}. {c}^{ - 8} )}^{ - 3} \\ = ({4}^{ - 3} . {a}^{ - 6 \times - 3} . {b}^{15 \times - 3} . {c}^{ - 8 \times - 3} ) \\ = ( {4}^{ - 3} . {a}^{18} . {b}^{ - 45} . {c}^{24} ) \\ = \frac{{a}^{18}.{c}^{24}}{64{b}^{ 45}}$

$(c) \: = {(3x \times 5y \times 8z)}^{ - 2} \\ = {(120xyz)}^{ - 2} \\ = {( \frac{1}{120xyz} )}^{2} \\ = \frac{1}{14400 {x}^{2} {y}^{2} {z}^{2} }$