`2^2-3^4+4^3-5^2+7^2=(?)+sqrt9`

Options

A. `(36)^2`
B. 8
C. 6
D. 36

Answer

To solve the equation \(2^2 - 3^4 + 4^3 - 5^2 + 7^2 = (?)+\sqrt{9}\), we will follow these steps:

### Step 1: Calculate each term on the left side of the equation.
- \(2^2 = 4\)
- \(3^4 = 81\)
- \(4^3 = 64\)
- \(5^2 = 25\)
- \(7^2 = 49\)

### Step 2: Substitute the calculated values back into the equation.
Now, substituting these values into the equation gives us:
\[ 4 - 81 + 64 - 25 + 49 \]

### Step 3: Perform the addition and subtraction in order.
1. Start with \(4 - 81\):
\[ 4 - 81 = -77 \]
2. Now add \(64\):
\[ -77 + 64 = -13 \]
3. Next, subtract \(25\):
\[ -13 - 25 = -38 \]
4. Finally, add \(49\):
\[ -38 + 49 = 11 \]

### Step 4: Set the left side equal to the right side.
Now we have:
\[ 11 = x + \sqrt{9} \]

### Step 5: Calculate \(\sqrt{9}\).
Since \(\sqrt{9} = 3\), we can substitute this back into the equation:
\[ 11 = x + 3 \]

### Step 6: Solve for \(x\).
To find \(x\), subtract \(3\) from both sides:
\[ x = 11 - 3 \]
\[ x = 8 \]

### Final Answer:
Thus, the value of \(x\) is \(8\).

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