When evaluating open paren 6 x plus 9 close paren squared minus 5 for x is equal to 1 , the given value of 1 is substituted for what part of the expression?

In the expression (6x + 9)² − 5, x is the variable. When x = 1, substitute 1 wherever x appears, giving (6 × 1 + 9)² − 5. Therefore, 1 is substituted for the variable, choice D.

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