which is the best estimate of $sqrt{47}$ to the nearest tenth?\n6.8\n6.9\n7.0\n7.1

Explanation:

Step1: Find perfect - square bounds

We know that $6^2 = 36$ and $7^2=49$. Since $47$ is closer to $49$ than to $36$.

Step2: Use linear approximation (optional but helpful)

Let $y = \sqrt{x}$, and consider the interval $[49,36]$. The derivative of $y=\sqrt{x}=x^{\frac{1}{2}}$, and $y^\prime=\frac{1}{2\sqrt{x}}$. Near $x = 49$, $y^\prime=\frac{1}{2\sqrt{49}}=\frac{1}{14}$. The change in $x$ is $\Delta x=49 - 47 = 2$. The change in $y$ is approximately $\Delta y=y^\prime\Delta x=\frac{1}{14}\times2=\frac{1}{7}\approx0.14$. Since $\sqrt{49}=7$, then $\sqrt{47}\approx7 - 0.14 = 6.86\approx6.9$.

Answer:

B. 6.9