Question

exercise 2

1. $3^2 = 3 \times 3$
2. $23 \div 0 = 0$
3. $2(1 - 2) = 4$
4. $|-10| = 10$
5. $-(46) = -46$
6. $(107 + 322) = (322 + 107)$
7. $\sqrt{121} = 11 \text{ or } -11$
8. $5 - 3 = -5 + 3$
9. $0 \div 82 = 0$
10. $x^3 = 3 \cdot x$

Explanation:

Step1: Verify exponent definition

$3^2 = 3 \times 3 = 9$, so this is True.

Step2: Check division by zero

Division by 0 is undefined, so $23 \div 0 = 0$ is False.

Step3: Simplify left-hand side

$2(1-2) = 2(-1) = -2
eq 4$, so this is False.

Step4: Evaluate absolute value

The absolute value of $-10$ is 10, so $|-10| = 10$ is True.

Step5: Simplify negative of positive

$-(46) = -46$, so this is True.

Step6: Apply commutative property

Addition is commutative, so $(107+322)=(322+107)$ is True.

Step7: Evaluate principal square root

The principal square root $\sqrt{121}$ is only 11 (not -11), so this is False.

Step8: Calculate both sides

$5-3=2$, $-5+3=-2$; $2
eq -2$, so this is False.

Step9: Check division of zero

$0 \div 82 = 0$, so this is True.

Step10: Simplify exponent expression

$x^3 = x \cdot x \cdot x
eq 3 \cdot x$ (for most $x$), so this is False.

Answer:

1. True
2. False
3. False
4. True
5. True
6. True
7. False
8. False
9. True
10. False