(n(1) + n(0))*20,65625

asked by guest
on Feb 01, 2025 at 5:14 am



MathBot Answer:

The sum of (n(1)+n(0))⋅20\left(n\left( 1 \right) + n\left( 0 \right)\right) \cdot 20 and 6562565625 is:

=((n(1)+n(0))⋅20)+(65625)=(n+0)20+65625=20n+65625\begin{aligned}&=\left(\left(n\left( 1 \right) + n\left( 0 \right)\right) \cdot 20\right) + \left(65625\right) \\\\ &= \left(n + 0\right) 20 + 65625 \\\\ &= 20 n + 65625 \\\\\end{aligned}

The difference of (n(1)+n(0))⋅20\left(n\left( 1 \right) + n\left( 0 \right)\right) \cdot 20 and 6562565625 is:

=((n(1)+n(0))⋅20)−(65625)=(n+0)20−65625=20n−65625\begin{aligned}&=\left(\left(n\left( 1 \right) + n\left( 0 \right)\right) \cdot 20\right) - \left(65625\right) \\\\ &= \left(n + 0\right) 20 - 65625 \\\\ &= 20 n - 65625 \\\\\end{aligned}

The product of (n(1)+n(0))⋅20\left(n\left( 1 \right) + n\left( 0 \right)\right) \cdot 20 and 6562565625 is:

=((n(1)+n(0))⋅20)⋅(65625)=1312500n\begin{aligned}&=\left(\left(n\left( 1 \right) + n\left( 0 \right)\right) \cdot 20\right) \cdot \left(65625\right) \\\\ &= 1312500 n \\\\\end{aligned}

The quotient of (n(1)+n(0))⋅20\left(n\left( 1 \right) + n\left( 0 \right)\right) \cdot 20 and 6562565625 is:

=((n(1)+n(0))⋅20)(65625)=4n13125\begin{aligned}&= \frac{\left(\left(n\left( 1 \right) + n\left( 0 \right)\right) \cdot 20\right)}{\left(65625\right)} \\\\ &= \frac{4 n}{13125} \\\\\end{aligned}

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