`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 defined operations for "@" and "#".

### Step 1: Evaluate `20"@"4`
According to the definition provided:
- \( x"@"y = (x + y)^2 \)

Substituting \( x = 20 \) and \( y = 4 \):
\[ 20"@"4 = (20 + 4)^2 \]
Calculating the sum:
\[ 20 + 4 = 24 \]
Now, squaring the result:
\[ (24)^2 = 576 \]

### Step 2: Evaluate `576#35`
Now we need to use the result from Step 1 in the next operation:
- \( x#y = (x - y)^2 \)

Substituting \( x = 576 \) and \( y = 35 \):
\[ 576#35 = (576 - 35)^2 \]
Calculating the difference:
\[ 576 - 35 = 541 \]
Now, squaring the result:
\[ (541)^2 \]

### Step 3: Calculate \( (541)^2 \)
To calculate \( 541^2 \):
\[ 541^2 = 541 \times 541 \]
Using multiplication:
\[ 541 \times 541 = 292681 \]

### Final Answer
Thus, the value of \( (20"@"4)#35 \) is:
\[ \boxed{292681} \]