Question

[2^-1+3^0+5^1+7^2+9^3]÷(2/3)^-1

Answer 1

Hey friend,
Here is the answer you were looking for:

(2^-1 + 3^0 + 5^1 +7^2 + 9^3) ÷ (2/3)^-1

= (1/2 + 1 + 5 + 49 + 729) ÷ (3/2)

= ( 1×1 + 1×2 + 5×2 + 49×2 + 729×2 / 2 ) ×(2/3)

= (1 + 2 + 10 + 98 × 1458 / 2) × (2/3)

= (1569/2) × (2/3)

= 1569/2 × 2/3

= 1569/3

= 523

Hope this helps!!!

@Mahak24

Thanks...
☺☺

Answer 2

$[ 2^{-1} + 3^0 + 5^1 + 7^2 + 9^3]$÷$(2/3)^{-1}$

$[ \frac{1}{2} + 1 + 5 + 49 + 729]$ ÷ $\frac{3}{2}$

$[ \frac{1}{2} + 784 ]$×$\frac{2}{3}$

$[\frac{1 +1568 }{2}]$×$\frac{2}{3}$

$\frac{1569}{2}$×$\frac{2}{3}$

$\frac{1569}{3}$


523 




i hope this will help you



-by ABHAY