approximate $sqrt{46}$ to the nearest hundred...

# Explanation: ## Step1: Find the two perfect - squares around 46 We know that $6^2 = 36$ and $7^2=49$. ## Step2: Use linear approximation Let $y = \sqrt{x}$, and we consider the interval $[36,49]$. The derivative of $y=\sqrt{x}=x^{\frac{1}{2}}$ is $y^\prime=\frac{1}{2\sqrt{x}}$. We use the formula for linear approximation $L(x)=f(a)+f^\prime(a)(x - a)$. Let $a = 49$ and $x = 46$. $f(x)=\sqrt{x}$, $f(49) = 7$, and $f^\prime(x)=\frac{1}{2\sqrt{x}}$, so $f^\prime(49)=\frac{1}{2\times7}=\frac{1}{14}$. $L(46)=7+\frac{1}{14}(46 - 49)=7-\frac{3}{14}\approx7 - 0.2143 = 6.7857$. ## Step3: Use a calculator Using a calculator, $\sqrt{46}\approx6.7823$. ## Step4: Round to the nearest hundredth Rounding 6.7823 to the nearest hundredth gives 6.78. # Answer: 6.78