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(C) quadratic equation is given by $ax^2 + bx + c = 0$,where $a 
eq 0$.Since $0$ is a solution,substituting $x = 0$ into the equation gives $a(0)^2 +

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$90$

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(C) quadratic equation is given by $ax^2 + bx + c = 0$,where $a 
eq 0$.Since $0$ is a solution,substituting $x = 0$ into the equation gives $a(0)^2 + b(0) + c = 0$,which implies $c = 0$.The coefficients $a, b, c$ are chosen from the set $\{0, 1, 2, \dots, 9\}$.For the equation to be quadratic,the coefficient $a$ must be non-zero,so $a \in \{1, 2, \dots, 9\}$. This gives $9$ possible choices for $a$.The coefficient $b$ can be any value from the set $\{0, 1, 2, \dots, 9\}$,which gives $10$ possible choices for $b$.The coefficient $c$ is fixed as $0$,so there is only $1$ choice for $c$.Therefore,the total number of such quadratic equations is $N = 9 	imes 10 	imes 1 = 90$.

Let $N$ be the number of quadratic equations of the form $ax^2 + bx + c = 0$ with coefficients $a, b, c \in \{0, 1, 2, \dots, 9\}$ such that $0$ is a solution of each equation. Then the value of $N$ is

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