A quantity P is partly constant and partly varies inversely as square of t. p=6 when t=6 and p=18 when t=3. Find t when p=11.

Tunajua $P=6$ wakati $t=6$:

$$6 = k_1 + \frac{k_2}{6^2}$$ $$6 = k_1 + \frac{k_2}{36} \quad \text{(Mlinganyo 1)}$$

Tunajua $P=18$ wakati $t=3$:

$$18 = k_1 + \frac{k_2}{3^2}$$ $$18 = k_1 + \frac{k_2}{9} \quad \text{(Mlinganyo 2)}$$

Toa Mlinganyo 1 kutoka Mlinganyo 2:

$$(18 - 6) = \left(k_1 + \frac{k_2}{9}\right) - \left(k_1 + \frac{k_2}{36}\right)$$ $$12 = \frac{k_2}{9} - \frac{k_2}{36}$$

Pata kigawo cha kawaida (LCM) cha 9 na 36, ambacho ni 36:

$$12 = \frac{4k_2 - k_2}{36}$$ $$12 = \frac{3k_2}{36}$$ $$12 = \frac{k_2}{12}$$

Kisha, $k_2 = 12 \times 12 = 144$.

Badilisha thamani ya $k_2$ katika Mlinganyo 1:

$$6 = k_1 + \frac{144}{36}$$ $$6 = k_1 + 4$$ $$k_1 = 6 - 4$$ $$k_1 = 2$$

Kwa hiyo, uhusiano kati ya P na t ni $P = 2 + \frac{144}{t^2}$.

Badilisha $P=11$ katika mlinganyo:

$$11 = 2 + \frac{144}{t^2}$$ $$11 - 2 = \frac{144}{t^2}$$ $$9 = \frac{144}{t^2}$$ $$t^2 = \frac{144}{9}$$ $$t^2 = 16$$ $$t = \pm\sqrt{16}$$

Kwa kuwa $t$ kwa kawaida ni thamani chanya katika muktadha huu:

$$t = 4$$ $$\boxed{t = 4}$$

9. Kokotoa thamani kamili ya $0.23 - 0.14$.

$$0.23 - 0.14 = 0.09$$ $$\boxed{0.09}$$