(n(1) + n(0)) *20.65625
asked by guest on Jan 28, 2025 at 3:14 pm
You asked:
MathBot Answer:
(n(1)+n(0))⋅20.65625=661n32\displaystyle \left(n\left( 1 \right) + n\left( 0 \right)\right) \cdot 20.65625 = \frac{661 n}{32} (n(1)+n(0))⋅20.65625=32661n
(n(1)+n(0))⋅20.65625=661n32\left(n\left( 1 \right) + n\left( 0 \right)\right) \cdot 20.65625 = \frac{661 n}{32}(n(1)+n(0))⋅20.65625=32661n