Question

-6+3\cdot5 blank \sqrt{16}+32\div8 which one is greater

Answer 1

so if im righjt it would be > symbol to show that 9 is greater

Answer 2

Let's solve both expressions step by step.

First, we'll simplify the expression on the left:

\[
-6 + 3 \cdot 5
\]

Calculating \(3 \cdot 5\):

\[
3 \cdot 5 = 15
\]

Now, add \(-6\) to \(15\):

\[
-6 + 15 = 9
\]

Next, let's simplify the expression on the right:

\[
\sqrt{16} + 32 \div 8
\]

Calculating \(\sqrt{16}\):

\[
\sqrt{16} = 4
\]

Now, calculate \(32 \div 8\):

\[
32 \div 8 = 4
\]

Combining these results:

\[
4 + 4 = 8
\]

Now we can compare the two results:

- Left side: \(9\)
- Right side: \(8\)

So, \(9\) (from the left expression) is greater than \(8\) (from the right expression).

Thus, the greater value is \(9\).

Answer 3

Yes, that's correct! You would use the "greater than" symbol (>) to show that \(9\) is greater than \(8\).

So you can express it as:

\[
9 > 8
\]

This indicates that \(9\) is greater than \(8\).