Question

estimate \\(\sqrt{50}\\) to the nearest tenth. then locate \\(\sqrt{50}\\) on a number line. \\(\sqrt{50} \approx \square\\) (round to the nearest tenth as needed.)

Explanation:

Step1: Identify nearby perfect squares

We know that $7^2 = 49$ and $8^2 = 64$. Since $49 < 50 < 64$, $\sqrt{50}$ is between 7 and 8.

Step2: Calculate decimal approximation

Calculate $7.1^2 = 50.41$ and $7.0^2 = 49$. Since $50$ is closer to $50.41$ than to $49$, we check the difference:
$50 - 49 = 1$, $50.41 - 50 = 0.41$.
Since $0.41 < 1$, $\sqrt{50}$ is closer to $7.1$ than to $7.0$.

Step3: Confirm rounding to nearest tenth

$\sqrt{50} \approx 7.1$ when rounded to the nearest tenth.
For the number line: Mark 7 and 8, divide the interval between them into 10 equal parts. $\sqrt{50}$ is located 1 small part to the right of 7 (at the position of 7.1).

Answer:

$\sqrt{50} \approx 7.1$
To locate on a number line: Place a point between 7 and 8, 1 tenth of the way from 7 toward 8.