A quantity P is partly constant and partly varies as the square root of a quantity Q. Given that P = 20 when Q = 4 and that P = 60 when Q = 100, find Q when P = 22.

Let P be partly constant $k_1$ and partly vary as $\sqrt{Q}$ with constant $k_2$.

$$P = k_1 + k_2\sqrt{Q}$$

When $P = 20$, $Q = 4$:

$$20 = k_1 + k_2\sqrt{4}$$ $$20 = k_1 + 2k_2 \quad \text{(Equation 1)}$$

When $P = 60$, $Q = 100$:

$$60 = k_1 + k_2\sqrt{100}$$ $$60 = k_1 + 10k_2 \quad \text{(Equation 2)}$$

Subtract Equation 1 from Equation 2:

$$(60 - 20) = (k_1 + 10k_2) - (k_1 + 2k_2)$$ $$40 = 8k_2$$ $$k_2 = \frac{40}{8} = 5$$

Substitute $k_2 = 5$ into Equation 1:

$$20 = k_1 + 2(5)$$ $$20 = k_1 + 10$$ $$k_1 = 20 - 10 = 10$$

$$22 = 10 + 5\sqrt{Q}$$ $$22 - 10 = 5\sqrt{Q}$$ $$12 = 5\sqrt{Q}$$ $$\sqrt{Q} = \frac{12}{5}$$

Square both sides:

$$Q = \left(\frac{12}{5}\right)^2$$ $$Q = \frac{144}{25}$$ $$Q = 5.76$$

When P = 22, Q is:

$$\boxed{5.76}$$

T varies directly as $(10-U)^2$ and inversely as $\sqrt[3]{V}$.

$$T = k \frac{(10-U)^2}{\sqrt[3]{V}}$$

where $k$ is the constant of proportionality.

Given $T = 12$, $U = 4$, $V = 8$.

$$12 = k \frac{(10-4)^2}{\sqrt[3]{8}}$$ $$12 = k \frac{(6)^2}{2}$$ $$12 = k \frac{36}{2}$$ $$12 = k \times 18$$ $$k = \frac{12}{18} = \frac{2}{3}$$

$$T = \frac{2}{3} \frac{(10-U)^2}{\sqrt[3]{V}}$$

The equation connecting T, U, and V is:

$$\boxed{T = \frac{2(10-U)^2}{3\sqrt[3]{V}}}$$

$$T = 5\frac{2}{5} = \frac{5 \times 5 + 2}{5} = \frac{27}{5}$$ $$V = 15\frac{5}{8} = \frac{15 \times 8 + 5}{8} = \frac{120 + 5}{8} = \frac{125}{8}$$

$$\frac{27}{5} = \frac{2}{3} \frac{(10-U)^2}{\frac{5}{2}}$$ $$\frac{27}{5} = \frac{2}{3} \times \frac{2}{5} (10-U)^2$$ $$\frac{27}{5} = \frac{4}{15} (10-U)^2$$

$$(10-U)^2 = \frac{27}{5} \times \frac{15}{4}$$ $$(10-U)^2 = \frac{27 \times 3}{4}$$ $$(10-U)^2 = \frac{81}{4}$$

Take the square root of both sides:

$$10-U = \pm\sqrt{\frac{81}{4}}$$ $$10-U = \pm\frac{9}{2}$$

Case 1: $10-U = \frac{9}{2}$

$$U = 10 - \frac{9}{2}$$ $$U = \frac{20}{2} - \frac{9}{2}$$ $$U = \frac{11}{2} = 5.5$$

Case 2: $10-U = -\frac{9}{2}$

$$U = 10 + \frac{9}{2}$$ $$U = \frac{20}{2} + \frac{9}{2}$$ $$U = \frac{29}{2} = 14.5$$

The possible values for U are:

$$\boxed{U = 5.5 \text{ or } U = 14.5}$$