Question

Evaluate : [(2⁰+ 3¹)x3²]÷[(3²-2³)x3³]​

Answer 1

$[(2^0+3^1)*3^2] / [(3^2-2^3)*3^3]$

*Evaluate the value of the exponents*

[(2⁰+ 3¹)x3²]

= (1+3) x 9

= 4 x 9

= 36

[(3²-2³)x3³]​

= (9-8) x 27

= 1 x 27

= 27

*Divide the values*

$[(2^0+3^1)*3^2] / [(3^2-2^3)*3^3]\\= 36 / 27\\= 1.3$

Therefore,

$[(2^0+3^1)*3^2] / [(3^2-2^3)*3^3]= 1.3$