How do you simplify $\frac { ( - 8) ^ { 4} \cdot 16^ { - 3} \cdot 35^ { 3} } { 14^ { 3} \cdot 50^ { 2} \cdot 24^ { - 2} }$ ?

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Answer 1 · 2018-05-26T04:04:51

$18/5 =3.6$

$((-8)^4*16^-3*35^3)/(14^3*50^2*24^-2$

= $(-8)^4*16^-3*35^3*14^-3*50^-2*24^2$

= $((-2)^3)^4*(2^4)^-3*5^3*7^3*2^-3*7^-3*2^-2*(5^2)^-2*3^2*(2^3)^2$

= $(-2)^12*2^-12*2^(-3-2+6)*3^2*5^(3-4)*7^(3-3)$

= $2^12*2^-12*2^1*3^2*5^-1*7^0$

= $(2*9*1)/5$

= $18/5$

= $3.6$

Note:
$1." "(-2)^12 = 2^12$ since exponent is even
$2." "7^0 = 1$ .... by definition

Answer 2 · 2018-05-26T09:02:24

$18/5$

$\frac { ( - 8) ^ { 4} \cdot 16^ { - 3} \cdot 35^ { 3} } { 14^ { 3} \cdot 50^ { 2} \cdot 24^ { - 2} }$

Write the bases as the product of their prime factors:

$((-2^3)^4 * (2^4)^-3*(5xx7)^3)/((2xx7)^3*(2xx5^2)^2* (2^3xx3)^-2$

Multiply the indices to remove the brackets.

$=(2^12 * 2^-12 * 5^3 * 7^3)/(2^3 * 7^3 * 2^2 * 5^4 * 2^-6 * 3^-2)$

Simplify by adding the indices of like bases

$= (2^0 * 5^3 * 7^3)/(2^-1 * 3^-2 * 5^4* 7^3)" "larr (7^3/7^3 =1)$

$=(2 * 3^2)/5" "larr" "$ law: $1/x^-m = x^m$

$=18/5$

How do you simplify \frac { ( - 8) ^ { 4} \cdot 16^ { - 3} \cdot 35^ { 3} } { 14^ { 3} \cdot 50^ { 2} \cdot 24^ { - 2} }\frac { ( - 8) ^ { 4} \cdot 16^ { - 3} \cdot 35^ { 3} } { 14^ { 3} \cdot 50^ { 2} \cdot 24^ { - 2} }?