Question

3^2 > 9 * ? = 20

Answer 1

its a whole number

Answer 2

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:

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

Answer 3

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 \( 9 * ? < 20 \) 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 \( 3^2 > 9 * ? \).