Simplify the following expressions and solve the inequalities: a) y2 y3, b) 50 6, c) (F6 F3) / (F4 F2), d) n7 / n2, e) 56 / 54, f) 9c5 / 9c2, and g) 1/4x - 4 <= 6.

$$y^2 \times y^3 = y^{2+3} = y^5$$

The simplified expression is $\boxed{y^5}$.

b) $5^0 \times 6$

$$5^0 \times 6 = 1 \times 6 = 6$$

The simplified expression is $\boxed{6}$.

c) $(F^6 \times F^3) \div (F^4 \times F^2)$

$$(F^6 \times F^3) = F^{6+3} = F^9$$ $$(F^4 \times F^2) = F^{4+2} = F^6$$

$$F^9 \div F^6 = F^{9-6} = F^3$$

The simplified expression is $\boxed{F^3}$.

d) $n^7 \div n^2$

$$n^7 \div n^2 = n^{7-2} = n^5$$

The simplified expression is $\boxed{n^5}$.

e) $5^6 \div 5^4$

$$5^2 = 25$$

The simplified expression is $\boxed{25}$.

f) $9c^5 \div 9c^2$

$$9c^5 \div 9c^2 = \frac{9}{9} \times \frac{c^5}{c^2} = 1 \times c^{5-2} = c^3$$

The simplified expression is $\boxed{c^3}$.

2. Solve the following inequalities

a) $\frac{1}{4}x - 4 \le 6$

$$\frac{1}{4}x - 4 + 4 \le 6 + 4$$ $$\frac{1}{4}x \le 10$$

$$4 \times \frac{1}{4}x \le 10 \times 4$$ $$x \le 40$$

The solution is $\boxed{x \le 40}$.

b) $3x + 5 > -4$

$$3x + 5 - 5 > -4 - 5$$ $$3x > -9$$

$$\frac{3x}{3} > \frac{-9}{3}$$ $$x > -3$$

The solution is $\boxed{x > -3}$.

c) $\frac{3 - 2x}{3} > -8\frac{1}{3}$

$$-8\frac{1}{3} = -\frac{(8 \times 3) + 1}{3} = -\frac{24+1}{3} = -\frac{25}{3}$$

The inequality becomes:

$$\frac{3 - 2x}{3} > -\frac{25}{3}$$

$$3 \times \frac{3 - 2x}{3} > 3 \times -\frac{25}{3}$$ $$3 - 2x > -25$$

$$3 - 2x - 3 > -25 - 3$$ $$-2x > -28$$

Remember to reverse the inequality sign when dividing by a negative number.

$$\frac{-2x}{-2} < \frac{-28}{-2}$$ $$x < 14$$

The solution is $\boxed{x < 14}$.

d) $5(2y+9) \ge 3(2y+9)$

$$10y - 6y + 45 \ge 6y - 6y + 27$$ $$4y + 45 \ge 27$$

$$4y + 45 - 45 \ge 27 - 45$$ $$4y \ge -18$$

$$\frac{4y}{4} \ge \frac{-18}{4}$$ $$y \ge -\frac{9}{2}$$ $$y \ge -4.5$$

The solution is $\boxed{y \ge -4.5}$.

3. Work out the x and y intercept for the line whose equations are

a) $y = 4x + 3$
To find the x-intercept, set $y=0$:

$$0 = 4x + 3$$ $$-3 = 4x$$ $$x = -\frac{3}{4}$$

To find the y-intercept, set $x=0$:

$$y = 4(0) + 3$$ $$y = 3$$

The x-intercept is $\boxed{\left(-\frac{3}{4}, 0\right)}$ and the y-intercept is $\boxed{(0, 3)}$.

b) $3x + 2y = 12$
To find the x-intercept, set $y=0$:

$$3x + 2(0) = 12$$ $$3x = 12$$ $$x = \frac{12}{3}$$ $$x = 4$$

To find the y-intercept, set $x=0$:

$$3(0) + 2y = 12$$ $$2y = 12$$ $$y = \frac{12}{2}$$ $$y = 6$$

The x-intercept is $\boxed{(4, 0)}$ and the y-intercept is $\boxed{(0, 6)}$.

c) $y = -2x + 3$
To find the x-intercept, set $y=0$:

$$0 = -2x + 3$$ $$2x = 3$$ $$x = \frac{3}{2}$$

To find the y-intercept, set $x=0$:

$$y = -2(0) + 3$$ $$y = 3$$

The x-intercept is $\boxed{\left(\frac{3}{2}, 0\right)}$ and the y-intercept is $\boxed{(0, 3)}$.

4. Find the volume and surface of the following

a) sphere radius 3.5 cm
Volume of a sphere: $V = \frac{4}{3}\pi r^3$
Surface area of a sphere: $SA = 4\pi r^2$
Given $r = 3.5 \text{ cm}$.

$$V = \frac{4}{3} \times \pi \times (3.5 \text{ cm})^3$$ $$V = \frac{4}{3} \times \pi \times 42.875 \text{ cm}^3$$ $$V \approx 179.59 \text{ cm}^3$$

$$SA = 4 \times \pi \times (3.5 \text{ cm})^2$$ $$SA = 4 \times \pi \times 12.25 \text{ cm}^2$$ $$SA \approx 153.94 \text{ cm}^2$$

The volume is $\boxed{179.59 \text{ cm}^3}$ and the surface area is $\boxed{153.94 \text{ cm}^2}$.

b) cone radius 7 cm and slant height 10 cm
Volume of a cone: $V = \frac{1}{3}\pi r^2 h$
Surface area of a cone: $SA = \pi r (r+l)$
Given $r = 7 \text{ cm}$ and $l = 10 \text{ cm}$.
First, find the height $h$ using the Pythagorean theorem: $h = \sqrt{l^2 - r^2}$.

$$h = \sqrt{(10 \text{ cm})^2 - (7 \text{ cm})^2}$$ $$h = \sqrt{100 \text{ cm}^2 - 49 \text{ cm}^2}$$ $$h = \sqrt{51} \text{ cm} \approx 7.14 \text{ cm}$$

$$V = \frac{1}{3} \times \pi \times (7 \text{ cm})^2 \times \sqrt{51} \text{ cm}$$ $$V = \frac{1}{3} \times \pi \times 49 \text{ cm}^2 \times 7.1414 \text{ cm}$$ $$V \approx 366.49 \text{ cm}^3$$

$$SA = \pi \times 7 \text{ cm} \times (7 \text{ cm} + 10 \text{ cm})$$ $$SA = \pi \times 7 \text{ cm} \times 17 \text{ cm}$$ $$SA = 119\pi \text{ cm}^2$$ $$SA \approx 373.85 \text{ cm}^2$$

The volume is $\boxed{366.49 \text{ cm}^3}$ and the surface area is $\boxed{373.85 \text{ cm}^2}$.

c) cuboid of 5 m x 2 m x 3.2 m
Volume of a cuboid: $V = lwh$
Surface area of a cuboid: $SA = 2(lw + lh + wh)$
Given $l = 5 \text{ m}$, $w = 2 \text{ m}$, $h = 3.2 \text{ m}$.

$$V = 5 \text{ m} \times 2 \text{ m} \times 3.2 \text{ m}$$ $$V = 32 \text{ m}^3$$

$$SA = 2((5 \text{ m} \times 2 \text{ m}) + (5 \text{ m} \times 3.2 \text{ m}) + (2 \text{ m} \times 3.2 \text{ m}))$$ $$SA = 2(10 \text{ m}^2 + 16 \text{ m}^2 + 6.4 \text{ m}^2)$$ $$SA = 2(32.4 \text{ m}^2)$$ $$SA = 64.8 \text{ m}^2$$

The volume is $\boxed{32 \text{ m}^3}$ and the surface area is $\boxed{64.8 \text{ m}^2}$.

d) cube of sides 4 m
Volume of a cube: $V = s^3$
Surface area of a cube: $SA = 6s^2$
Given $s = 4 \text{ m}$.

$$V = (4 \text{ m})^3$$ $$V = 64 \text{ m}^3$$

$$SA = 6 \times (4 \text{ m})^2$$ $$SA = 6 \times 16 \text{ m}^2$$ $$SA = 96 \text{ m}^2$$

The volume is $\boxed{64 \text{ m}^3}$ and the surface area is $\boxed{96 \text{ m}^2}$.

5. A car is brought to rest from 90 km/h in 20 seconds. What is its acceleration?
Given:
Initial velocity, $v_i = 90 \text{ km/h}$
Final velocity, $v_f = 0 \text{ km/h}$ (brought to rest)
Time, $t = 20 \text{ s}$