\(4.8^{2}+28.6^{2}=x^{2}\) round to the nearest tenth if necessary

Answer 1

To find \( x \) in the equation \( 4.8^2 + 28.6^2 = x^2 \), we first need to calculate \( 4.8^2 \) and \( 28.6^2 \).

1. Calculate \( 4.8^2 \): \[ 4.8^2 = 23.04 \]

2. Calculate \( 28.6^2 \): \[ 28.6^2 = 817.96 \]

3. Now, add both results: \[ 4.8^2 + 28.6^2 = 23.04 + 817.96 = 841.00 \]

4. Now, take the square root to solve for \( x \): \[ x = \sqrt{841} = 29 \]

Thus, the value of \( x \) rounded to the nearest tenth is: \[ \boxed{29.0} \]