Question: User: 40-[3^2\times 3+2\times 2^3\times 2-(-10)\times 2]+90\div(-9)=
1. **State the problem:** Simplify the expression $$40-[3^2\times 3+2\times 2^3\times 2-(-10)\times 2]+90\div(-9)$$.
2. **Recall order of operations:** Use PEMDAS (Parentheses, Exponents, Multiplication and Division from left to right, Addition and Subtraction from left to right).
3. **Calculate exponents:**
$$3^2 = 9$$
$$2^3 = 8$$
4. **Substitute exponents back:**
$$40-[9\times 3 + 2 \times 8 \times 2 - (-10) \times 2] + 90 \div (-9)$$
5. **Perform multiplications inside the brackets:**
$$9 \times 3 = 27$$
$$2 \times 8 \times 2 = 32$$
$$-(-10) \times 2 = 20$$
6. **Substitute these values:**
$$40 - [27 + 32 + 20] + 90 \div (-9)$$
7. **Add inside the brackets:**
$$27 + 32 + 20 = 79$$
8. **Simplify brackets:**
$$40 - 79 + 90 \div (-9)$$
9. **Perform division:**
$$90 \div (-9) = -10$$
10. **Substitute division result:**
$$40 - 79 - 10$$
11. **Perform subtraction from left to right:**
$$40 - 79 = -39$$
$$-39 - 10 = -49$$
**Final answer:** $$-49$$