Question

which number is closest to \\(\sqrt{84}\\) ? 9.8 10.1 9.1 8.7

Explanation:

Step1: Calculate $\sqrt{84}$

We know that $9^2 = 81$ and $10^2 = 100$. So $\sqrt{84}$ is between $9$ and $10$. Calculate $\sqrt{84}\approx9.165$.

Step2: Find the difference from each option

- For $9.8$: $|9.8 - 9.165| = 0.635$
- For $10.1$: $|10.1 - 9.165| = 0.935$
- For $9.1$: $|9.1 - 9.165| = 0.065$
- For $8.7$: $|8.7 - 9.165| = 0.465$

Step3: Compare differences

The smallest difference is $0.065$ for $9.1$? Wait, no, wait, recalculate $\sqrt{84}$ more accurately. Wait, $9.1^2 = 82.81$, $9.2^2 = 84.64$. So $\sqrt{84}$ is between $9.1$ and $9.2$. Wait, $9.1^2 = 82.81$, $9.16^2=(9 + 0.16)^2 = 81 + 2.88 + 0.0256 = 83.9056$, $9.17^2=(9.16 + 0.01)^2 = 9.16^2 + 2\times9.16\times0.01 + 0.01^2 = 83.9056 + 0.1832 + 0.0001 = 84.0889$. So $\sqrt{84}\approx9.165$. Now check the differences again:
- $|9.8 - 9.165| = 0.635$
- $|10.1 - 9.165| = 0.935$
- $|9.1 - 9.165| = 0.065$
- $|8.7 - 9.165| = 0.465$
Wait, but wait, maybe I made a mistake. Wait, $9.1^2 = 82.81$, $9.2^2 = 84.64$. So $84$ is between $9.1^2$ and $9.2^2$. The distance from $84$ to $81$ (which is $9^2$) is $3$, and to $100$ (which is $10^2$) is $16$. Wait, no, better to calculate the square of each option:
- $9.8^2 = 96.04$
- $10.1^2 = 102.01$
- $9.1^2 = 82.81$
- $8.7^2 = 75.69$
Now find the difference between $84$ and each square:
- $|84 - 96.04| = 12.04$
- $|84 - 102.01| = 18.01$
- $|84 - 82.81| = 1.19$
- $|84 - 75.69| = 8.31$
Wait, no, that's not the right way. We need to find which number's square is closest to $84$. Wait, $9.1^2 = 82.81$, $9.2^2 = 84.64$. The difference between $84$ and $82.81$ is $1.19$, between $84$ and $84.64$ is $0.64$. Wait, but the options are $9.8$, $10.1$, $9.1$, $8.7$. Wait, maybe I miscalculated $\sqrt{84}$. Let's use a calculator: $\sqrt{84}\approx9.16515138991$. Now calculate the absolute difference between this value and each option:
- $|9.16515 - 9.8| = 0.63485$
- $|9.16515 - 10.1| = 0.93485$
- $|9.16515 - 9.1| = 0.06515$
- $|9.16515 - 8.7| = 0.46515$
Wait, but the smallest difference is with $9.1$? But wait, $9.1^2 = 82.81$, which is $1.19$ less than $84$, while $9.8^2 = 96.04$, which is $12.04$ more, $10.1^2 = 102.01$, $18.01$ more, $8.7^2 = 75.69$, $8.31$ less. Wait, no, I think I messed up the approach. The question is which number is closest to $\sqrt{84}$, not which number's square is closest to $84$. So we should calculate the distance between $\sqrt{84}\approx9.165$ and each option:
- $9.8 - 9.165 = 0.635$
- $10.1 - 9.165 = 0.935$
- $9.165 - 9.1 = 0.065$
- $9.165 - 8.7 = 0.465$
So the smallest distance is $0.065$ with $9.1$? But wait, let's check again. Wait, maybe I made a mistake in calculating $\sqrt{84}$. Wait, $9^2 = 81$, $10^2 = 100$. $84 - 81 = 3$, $100 - 84 = 16$. So $\sqrt{84}$ is closer to $9$ than to $10$. Now, $9.1^2 = 82.81$, $9.2^2 = 84.64$. So $84$ is $84 - 82.81 = 1.19$ above $9.1^2$, and $84.64 - 84 = 0.64$ below $9.2^2$. So $\sqrt{84}$ is closer to $9.2$ than to $9.1$, but $9.2$ is not an option. The options are $9.8$, $10.1$, $9.1$, $8.7$. Wait, maybe the question is which number is closest, so among the options, $9.1$ has a square of $82.81$, which is $1.19$ away from $84$, $9.8$ has a square of $96.04$, which is $12.04$ away, $10.1$ has $102.01$, $18.01$ away, $8.7$ has $75.69$, $8.31$ away. Wait, no, that's not the right way. The correct way is to find the number whose value is closest to $\sqrt{84}\approx9.165$. So:
- $9.8 - 9.165 = 0.635$
- $10.1 - 9.165 = 0.935$
- $9.165 - 9.1 = 0.065$
- $9.165 - 8.7 = 0.465$
So the smallest difference is $0.065$ with $9.1$. But wait, let's check with a calculator. Let's compute $\sqrt{84}\approx9.16515$. Now, $9.1$ is $9.1$, difference is $0.06515$. $9.8$ is $9.8$, difference is $0.63485$. $10.1$ is $10.1$, difference is $0.93485$. $8.7$ is $8.7$, difference is $0.46515$. So yes, $9.1$ is the closest? Wait, but wait, maybe I made a mistake. Wait, $9.1^2 = 82.81$, $9.2^2 = 84.64$. So $84$ is $84 - 82.81 = 1.19$ more than $9.1^2$, and $84.64 - 84 = 0.64$ less than $9.2^2$. So $\sqrt{84}$ is approximately $9.1 + \frac{3}{1.83}\approx9.1 + 1.64\approx9.26$? Wait, no, the linear approximation: let $f(x) = \sqrt{x}$, $f(81) = 9$, $f'(x) = \frac{1}{2\sqrt{x}}$, so $f(84) \approx f(81) + f'(81)(84 - 81) = 9 + \frac{1}{18}\times3 = 9 + \frac{1}{6}\approx9.1667$, which matches the earlier calculation. So $\sqrt{84}\approx9.1667$. Now, the difference between $9.1667$ and $9.1$ is $0.0667$, between $9.1667$ and $9.8$ is $0.6333$, between $9.1667$ and $10.1$ is $0.9333$, between $9.1667$ and $8.7$ is $0.4667$. So the smallest difference is with $9.1$. Wait, but the options are $9.8$, $10.1$, $9.1$, $8.7$. So the closest is $9.1$? But wait, maybe I made a mistake. Wait, let's check the square of $9.1$: $9.1\times9.1 = 82.81$, square of $9.2$ is $84.64$, so $84$ is $84 - 82.81 = 1.19$ above $9.1^2$, and $84.64 - 84 = 0.64$ below $9.2^2$. So $\sqrt{84}$ is closer to $9.2$ than to $9.1$, but $9.2$ is not an option. Among the given options, $9.1$ is the closest. Wait, but let's check the options again. The options are $9.8$, $10.1$, $9.1$, $8.7$. So the answer should be $9.1$? Wait, no, wait, maybe I messed up. Wait, $9.8^2 = 96.04$, which is way higher, $10.1^2 = 102.01$, also higher, $8.7^2 = 75.69$, lower, $9.1^2 = 82.81$, lower. So the closest in value to $\sqrt{84}\approx9.165$ is $9.1$, because the difference is about $0.065$, while the others are larger. So the answer is $9.1$.

Answer:

9.1