netres in scientific notation. gine you could take a piece of string and tie it around the Equator. Imagine increasing the th of this piece of string by 36 cm and arranging it so that it stands out evenly around the tor. How far do you think the string would be from the surface of the Earth? f the radius of the Earth is \( r \) and the distance from the string to the surface of he Earth is \( x \), prove that the distance is approximately 6 cm .

To solve this problem, we need to establish a relationship between the radius of the Earth, the length of the string, and the distance from the string to the surface of the Earth. ### Known Conditions: 1. The radius of the Earth is denoted as \( r \). 2. The original circumference of the Earth is given by the formula \( C = 2\pi r \). 3. The string is increased by 36 cm, so the new length of the string is \( C + 36 \) cm. 4. We need to find the distance \( x \) from the string to the surface of the Earth. ### Step-by-Step Solution: 1. **Calculate the new circumference of the string:** \[ C_{\text{new}} = C + 36 = 2\pi r + 36 \] 2. **Set the new circumference equal to the formula for circumference with the new radius:** The new radius \( r_{\text{new}} \) can be expressed in terms of the distance \( x \): \[ C_{\text{new}} = 2\pi (r + x) \] 3. **Equate the two expressions for the new circumference:** \[ 2\pi r + 36 = 2\pi (r + x) \] 4. **Simplify the equation:** \[ 2\pi r + 36 = 2\pi r + 2\pi x \] Subtract \( 2\pi r \) from both sides: \[ 36 = 2\pi x \] 5. **Solve for \( x \):** \[ x = \frac{36}{2\pi} = \frac{18}{\pi} \] 6. **Calculate the numerical value of \( x \):** Using the approximation \( \pi \approx 3.14 \): \[ x \approx \frac{18}{3.14} \approx 5.73 \text{ cm} \] ### Conclusion: The distance \( x \) from the string to the surface of the Earth is approximately 5.73 cm, which can be rounded to about 6 cm. Thus, we have shown that if the radius of the Earth is \( r \) and the string is increased by 36 cm, the distance from the string to the surface of the Earth is approximately 6 cm.