Question

question 1 without using a calculator, find the value of $(2^2 \times 5)^{-2} - (2^{-3} \times 3)^2 + 261(40)^{-2}$. give your answer as a fraction without spaces.

Explanation:

Step1: Simplify \((2^{2} \times 5)^{-2}\)

First, calculate the exponent inside the parentheses: \(2^{2}=4\), so \(2^{2} \times 5 = 4\times5 = 20\). Then apply the negative exponent rule \(a^{-n}=\frac{1}{a^{n}}\), so \((20)^{-2}=\frac{1}{20^{2}}=\frac{1}{400}\).

Step2: Simplify \((2^{-3} \times 3)^{2}\)

Calculate \(2^{-3}=\frac{1}{2^{3}}=\frac{1}{8}\), then \(2^{-3} \times 3=\frac{1}{8}\times3=\frac{3}{8}\). Now square this result: \((\frac{3}{8})^{2}=\frac{3^{2}}{8^{2}}=\frac{9}{64}\).

Step3: Simplify \(261(40)^{-2}\)

First, \(40^{-2}=\frac{1}{40^{2}}=\frac{1}{1600}\). Then multiply by 261: \(261\times\frac{1}{1600}=\frac{261}{1600}\).

Step4: Combine all terms

Now we have the expression: \(\frac{1}{400}-\frac{9}{64}+\frac{261}{1600}\). First, find a common denominator, which is 1600. Convert each fraction: \(\frac{1}{400}=\frac{4}{1600}\), \(\frac{9}{64}=\frac{225}{1600}\). Now substitute back: \(\frac{4}{1600}-\frac{225}{1600}+\frac{261}{1600}=\frac{4 - 225+261}{1600}=\frac{(4 + 261)-225}{1600}=\frac{265 - 225}{1600}=\frac{40}{1600}=\frac{1}{40}\).

Answer:

\(\frac{1}{40}\)