Problem 1: Find the value of $$3 x - 2$$ when $$x = 5$$.
Solution: Substituting $$x = 5$$ into the expression $$3 x - 2$$, we get:
$$3 (5) - 2 = 15 - 2 = 13$$
Therefore, the value of $$3 x - 2$$ when $$x = 5$$ is 13.
Problem 2: Solve the equation $$2x + 5 = 15.$$
Solution: Subtracting 5 from both sides of the equation, we get:
$$2x + 5 - 5 = 15 - 5$$
$$2x = 10$$
Dividing both sides by 2, we get:
$$\frac{2x}{2} = \frac{10}{2}$$
$$x = 5$$
Therefore, the solution to the equation $$2x + 5 = 15$$ is $$x = 5$$.
Problem 3: Simplify the expression $$3(2x + 5) - 2(4x - 1).$$
Solution: Distributing the coefficients, we get:
$$6x + 15 - 8x + 2 = -2x + 17$$
Combining like terms, we get:
$$-2x + 6x + 15 + 2 = 17$$
$$4x + 17 = 17$$
Subtracting 17 from both sides, we get:
$$4x + 17 - 17 = 17 - 17$$
$$4x = 0$$
Dividing both sides by 4, we get:
$$\frac{4x}{4} = \frac{0}{4}$$
$$x = 0$$
Therefore, the simplified value of the expression $$3(2x + 5) - 2(4x - 1)$$ is 0.
Problem 4: Find the area of a rectangle with length $$l = 8$$ cm and width $$w = 5$$ cm.
Solution: The area of a rectangle is given by the formula:
$$A = l \times w$$
Substituting the given values into the formula, we get:
$$A = 8 \times 5 = 40$$ cm2
Therefore, the area of the rectangle is 40 cm2.
Problem 5: Find the volume of a cube with side length $$a = 4$$ cm.
Solution: The volume of a cube is given by the formula:
$$V = a^3$$
Substituting the given value into the formula, we get:
$$V = 4^3 = 64$$ cm3
Therefore, the volume of the cube is 64 cm3.
Problem 6: Simplify the expression $$(2x^2 + 3x - 1) - (x^2 - 2x - 5).$$
Solution: Distributing the parentheses, we get:
$$2x^2 + 3x - 1 - x^2 + 2x + 5 = x^2 + 5x + 4$$
Combining like terms, we get:
$$2x^2 - x^2 + 3x + 2x - 1 + 5 = x^2 + 5x + 4$$
$$x^2 + 5x + 4 = x^2 + 5x + 4$$
Therefore, the expression simplifies to $$x^2 + 5x + 4$$, and there are no further simplifications possible.
Problem 7: Find the value of $$x^2 - 2x + 1$$ when $$x = 3$$.
Solution: Substituting $$x = 3$$ into the expression $$x^2 - 2x + 1$$, we get:
$$(3)^2 - 2(3) + 1 = 9 - 6 + 1 = 4$$
Therefore, the value of $$x^2 - 2x + 1$$ when $$x = 3$$ is 4.
Problem 8: Solve the inequation $$2x - 5 > 11$$.
Solution: Adding 5 to both sides of the inequation, we get:
$$2x - 5 + 5 > 11 + 5$$
$$2x > 16$$
Dividing both sides by 2, we get:
$$\frac{2x}{2} > \frac{16}{2}$$
$$x > 8$$
Therefore, the solution to the inequation $$2x - 5 > 11$$ is $$x > 8$$.
Problem 9: Find the probability of getting a head when a fair coin is tossed.
Solution: When a fair coin is tossed, there are two possible outcomes: head or tail. Each outcome is equally likely, so the probability of getting a head is:
$$P(head) = \frac{1}{2}$$
Therefore, the probability of getting a head when a fair coin is tossed is 1/2.