Question

3(-2025)^0-4^-2+2^-3

Answer 1

Let's break down the expression step by step:

1. **Calculate \((-2025)^0\)**:
Any non-zero number raised to the power of 0 is 1.
\[
(-2025)^0 = 1
\]

2. **Calculate \(-4^{-2}\)**:
\[
-4^{-2} = -\frac{1}{4^2} = -\frac{1}{16}
\]

3. **Calculate \(2^{-3}\)**:
\[
2^{-3} = \frac{1}{2^3} = \frac{1}{8}
\]

Now substituting these values back into the expression:

\[
3(-2025)^0 - 4^{-2} + 2^{-3} = 3(1) - \left(-\frac{1}{16}\right) + \frac{1}{8}
\]

This simplifies to:

\[
3 + \frac{1}{16} + \frac{1}{8}
\]

To add these fractions, we need a common denominator. The smallest common denominator for 16 and 8 is 16.

Convert \(\frac{1}{8}\) to have a denominator of 16:

\[
\frac{1}{8} = \frac{2}{16}
\]

Now, we can add the fractions:

\[
3 + \frac{1}{16} + \frac{2}{16} = 3 + \frac{3}{16}
\]

Expressing \(3\) as a fraction with the same denominator:

\[
3 = \frac{48}{16}
\]

Now adding everything:

\[
\frac{48}{16} + \frac{3}{16} = \frac{51}{16}
\]

Thus, the final answer is:

\[
\frac{51}{16}
\]