If we need \( ? \) to be a whole number in the equation \( 9 * ? = 20 \), we can determine if \( 20 \) is divisible by \( 9 \): 1. Calculate the largest integer \( ? \) such that \( 9 * ? < 9 \): - \( ? = 0 \) gives \( 9 * 0 = 0 \) which is less than \( 9 \) - \( ? = 1 \) gives \( 9 * 1 = 9 \) which is still less than \( 9 \) - \( ? = 2 \) gives \( 9 * 2 = 18 \) which is still less than \( 9 \) - \( ? = 3 \) gives \( 9 * 3 = 27 \) which is greater than \( 9 \) The highest whole number for \( ? \) such that \( 9 * ? 9 * ? \).

3^2 > 9 * ? = 20

To solve the equation \( 3^2 > 9 * ? = 20 \), we first evaluate \( 3^2 \):

\[ 3^2 = 9 \]

Now, we rewrite the inequality:

\[ 9 > 9 * ? \]

We need to solve for \( ? \) in the equation \( 9 * ? = 20 \):

\[ ? = \frac{20}{9} \]

Now we can evaluate the inequality \( 9 > 9 * ? \):

\[ 9 > 9 * \left(\frac{20}{9}\right) \] \[ 9 > 20 \]

This statement is false. Thus, \( 3^2 > 9 * ? \) is incorrect for \( ? = \frac{20}{9} \).

So the solution to your question is:

\[ ? = \frac{20}{9} \]

its a whole number

If we need \( ? \) to be a whole number in the equation \( 9 * ? = 20 \), we can determine if \( 20 \) is divisible by \( 9 \):

Calculate the largest integer \( ? \) such that \( 9 * ? < 9 \):
- \( ? = 0 \) gives \( 9 * 0 = 0 \) which is less than \( 9 \)
- \( ? = 1 \) gives \( 9 * 1 = 9 \) which is still less than \( 9 \)
- \( ? = 2 \) gives \( 9 * 2 = 18 \) which is still less than \( 9 \)
- \( ? = 3 \) gives \( 9 * 3 = 27 \) which is greater than \( 9 \)

The highest whole number for \( ? \) such that \( 9 * ? < 20 \) is \( 2 \).

Therefore, the whole number \( ? \) that satisfies \( 9 * ? < 20 \) is \( 2 \).

In conclusion: \[ ? = 2 \]

So \( 9 * 2 = 18 \), and \( 3^2 = 9 \) which is indeed greater than \( 18 \), satisfying the inequality \( 3^2 > 9 * ? \).