Question

Find the value of
\[ \frac{(0.6)^4-(0.5)^4}{(0.6)^2+(0.5)^2} \]

• A. \(0.0121\)
• B. \(0.011\)
• C. \(0.11\)
• D. \(1.1\)

Hint:

Use \(a^4-b^4=(a^2-b^2)(a^2+b^2)\) to cancel the denominator.

Answer 1

The Correct Option is C

Step 1: Compute the squares first.
$0.6^2 = 0.36$ and $0.5^2 = 0.25$.

Step 2: Compute the fourth powers by squaring these results.
$0.6^4 = (0.6^2)^2 = 0.36^2 = 0.1296$
$0.5^4 = (0.5^2)^2 = 0.25^2 = 0.0625$

Step 3: Work out the numerator and denominator separately.
Numerator: $0.1296 - 0.0625 = 0.0671$
Denominator: $0.36 + 0.25 = 0.61$

Step 4: Divide to get the final value.
\[ \frac{0.0671}{0.61} = 0.11 \]
This matches exactly since $0.0671 = 0.11 \times 0.61$, confirming there is no rounding involved, the division comes out clean.

Final Answer:
Computing every power directly also gives 0.11, the same as the shortcut using the difference of fourth powers identity. \[ \boxed{0.11} \]