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

Answer 1

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 2

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

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 \(8\).