To find the whole number nearest to $\sqrt{97},$ we find the two whole numbers whose squares bracket $97.$ Since $9^2=81$ and $10^2=100,$ we have $9^2<97<10^2,$ so the whole number nearest to $\sqrt{97}$ is $\boxed{10}.$
To find the rational number to the nearest half, we find the two multiples of $1/2$ that bracket $\sqrt{97}.$ Since $(9+1/2)^2=91+9\cdot1+1/4=100\frac{1}{4}$ and $(10-1/2)^2=100-10+\frac{1}{4}=100-\frac{1}{4},$ we have $(9+1/2)^2>\sqrt{97}>(10-1/2)^2,$ so the rational number $\sqrt{97}$ is between $\boxed{\frac{9}{2}}$ and $\boxed{\frac{19}{2}}.$
Using a number line what whole number and rational number to the nearest half is 97−−√ between?
5 answers
can you remove the "{" i cant understand what the answer is with it
Certainly! The whole number nearest to $\sqrt{97}$ is 10.
To find the rational number to the nearest half, we find the two multiples of $1/2$ that bracket $\sqrt{97}.$ Since $(9+1/2)^2=91+9\cdot1+1/4=100\frac{1}{4}$ and $(10-1/2)^2=100-10+\frac{1}{4}=100-\frac{1}{4},$ we have $(9+1/2)^2>\sqrt{97}>(10-1/2)^2,$ so the rational number $\sqrt{97}$ is between $\frac{9}{2}$ and $\frac{19}{2}.$
To find the rational number to the nearest half, we find the two multiples of $1/2$ that bracket $\sqrt{97}.$ Since $(9+1/2)^2=91+9\cdot1+1/4=100\frac{1}{4}$ and $(10-1/2)^2=100-10+\frac{1}{4}=100-\frac{1}{4},$ we have $(9+1/2)^2>\sqrt{97}>(10-1/2)^2,$ so the rational number $\sqrt{97}$ is between $\frac{9}{2}$ and $\frac{19}{2}.$
wrong
Apologies for the confusion. Let's correct the solution.
To find the whole number nearest to $\sqrt{97},$ we find the two whole numbers whose squares bracket $97.$ Since $9^2=81$ and $10^2=100,$ we have $81 < 97 < 100.$ The whole number nearest to $\sqrt{97}$ is $\boxed{9}.$
To find the rational number to the nearest half, we need to consider the number line with intervals of $1/2.$ The closest multiples of $1/2$ to $\sqrt{97}$ are $9\frac{1}{2}$ and $10\frac{1}{2}.$ Since $9^2 = 81 < 97 < 100 = 10^2,$ we have $9 < \sqrt{97} < 10.$ Therefore, $\sqrt{97}$ is between $\boxed{9\frac{1}{2}}$ and $\boxed{10\frac{1}{2}}$ on the number line.
To find the whole number nearest to $\sqrt{97},$ we find the two whole numbers whose squares bracket $97.$ Since $9^2=81$ and $10^2=100,$ we have $81 < 97 < 100.$ The whole number nearest to $\sqrt{97}$ is $\boxed{9}.$
To find the rational number to the nearest half, we need to consider the number line with intervals of $1/2.$ The closest multiples of $1/2$ to $\sqrt{97}$ are $9\frac{1}{2}$ and $10\frac{1}{2}.$ Since $9^2 = 81 < 97 < 100 = 10^2,$ we have $9 < \sqrt{97} < 10.$ Therefore, $\sqrt{97}$ is between $\boxed{9\frac{1}{2}}$ and $\boxed{10\frac{1}{2}}$ on the number line.