Question

can anyone solve this
(2^-1 + 3^0 + 5^1 +9^3 )÷ (2/3)^-1​

Answer 1

$(2 {}^{ - 1} + 3 {}^{0} + 5 {}^{1} + 9 {}^{3} ) \div (\frac{2}{3} ) {}^{ - 1} \\ = ( \frac{1}{2} + 1 + 5 + 27) \div \frac{3}{2} \\ = ( \frac{1}{2} + 33) \times \frac{2}{3} \\ = \frac{67}{2} \times \frac{2}{3} \\ = \frac{67}{3}$

Answer 2

Answer:

Answer:

(2^-1+3^0+5^1+9^3)÷(2/3)^-1

=(1/2+1+5+729)÷1/(2/3)

=1471/2÷3/2

=1471/3

=490.33