Which of the following is an integer? Choices: (A)sqrt 6 (B)𝜋 (C)sqrt 2 (D)6

Evaluate choices for integers: We need to evaluate each choice to determine if it is an integer.Choice (A) - sqrt(6): Choice (A) is sqrt(6). The square root of 6 is not an integer because 6 is not a perfect square (the perfect squares closest to 6 are 4 and 9, which are the squares of 2 and 3, respectively).Choice (B) - pi: Choice (B) is π (pi). Pi is a well-known mathematical constant that represents the ratio of the circumference of a circle to its diameter. Pi is an irrational number, not an integer.Choice (C) - sqrt(2): Choice (C) is sqrt(2). The square root of 2 is not an integer because 2 is not a perfect square (the perfect squares closest to 2 are 1 and 4, which are the squares of 1 and 2, respectively).Choice (D) - 6: Choice (D) is 6. The number 6 is an integer because it is a whole number that can be written without a fractional or decimal component.

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Which of the following is an integer? $\newline$ Choices: $\newline$ (A) $\sqrt{6}$ $\newline$ (B) $\pi$ $\newline$ (C) $\sqrt{2}$ $\newline$ (D) $6$

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Algebra 2

Simplify radical expressions with root inside the root

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Q. Which of the following is an integer?

\newline

Choices:

\newline

(A)

\sqrt{6}

\newline

(B)

\pi

\newline

(C)

\sqrt{2}

\newline

(D)

6

Evaluate choices for integers: We need to evaluate each choice to determine if it is an integer.
Choice (A) - $\sqrt{6}$ : Choice (A) is $\sqrt{6}$ . The square root of $6$ is not an integer because $6$ is not a perfect square (the perfect squares closest to $6$ are $4$ and $9$ , which are the squares of $2$ and $3$ , respectively).
Choice (B) - $\pi$ : Choice (B) is $\pi$ (pi). Pi is a well-known mathematical constant that represents the ratio of the circumference of a circle to its diameter. Pi is an irrational number, not an integer.
Choice (C) - $\sqrt{2}$ : Choice (C) is $\sqrt{2}$ . The square root of $2$ is not an integer because $2$ is not a perfect square (the perfect squares closest to $2$ are $1$ and $4$ , which are the squares of $1$ and $2$ , respectively).
Choice (D) - $6$ : Choice (D) is $6$ . The number $6$ is an integer because it is a whole number that can be written without a fractional or decimal component.