x@y = `(x + y)^2`
x#y = `(x-y)^2`
Find the value of (20@4)#35:

Options

A. 280081
B. 292681
C. 262781
D. 301756

Answer

To solve the problem step by step, we need to evaluate the expression \((20@4) \# 35\) using the definitions of the operations given:

1. **Understanding the operations**:
- \(x@y = (x + y)^2\)
- \(x\#y = (x - y)^2\)

2. **Step 1: Calculate \(20@4\)**:
- Using the definition of \(x@y\):
\[ 20@4 = (20 + 4)^2 \]
- Calculate \(20 + 4\):
\[ 20 + 4 = 24 \]
- Now, square \(24\):
\[ 24^2 = 576 \]
- Therefore, \(20@4 = 576\).

3. **Step 2: Calculate \((20@4) \# 35\)**:
- Now we need to evaluate \(576 \# 35\) using the definition of \(x\#y\):
\[ 576 \# 35 = (576 - 35)^2 \]
- Calculate \(576 - 35\):
\[ 576 - 35 = 541 \]
- Now, square \(541\):
\[ 541^2 = 292681 \]

4. **Final Result**:
- Therefore, the value of \((20@4) \# 35\) is \(292681\).

### Summary of the Steps:
1. Calculate \(20 + 4\) and square the result to find \(20@4\).
2. Subtract \(35\) from the result of \(20@4\) and square the result to find \((20@4) \# 35\).