Solve for x and y. (2x + 2) : (y − 5) = 1 : 2

P inatofautiana moja kwa moja na mraba wa Q na kinyume na R.

$$P = k \frac{Q^2}{R}$$

$$2 = k \frac{4^2}{6}$$ $$2 = k \frac{16}{6}$$ $$2 = k \frac{8}{3}$$ $$k = \frac{2 \times 3}{8} = \frac{6}{8} = \frac{3}{4}$$

$$P = \frac{3}{4} \frac{8^2}{4}$$ $$P = \frac{3}{4} \frac{64}{4}$$ $$P = \frac{3}{4} \times 16$$ $$P = 3 \times 4$$ $$P = 12$$

Thamani ya P ni $\boxed{12}$.

Tuma swali linalofuata 📸

B inatofautiana kwa sehemu na mraba wa M na kwa sehemu na kinyume cha N.

$$B = k_1 M^2 + k_2 \frac{1}{N}$$

$$96 = k_1 (2)^2 + k_2 \frac{1}{\frac{1}{2}}$$ $$96 = 4k_1 + 2k_2 \quad \text{(Mlinganyo 1)}$$

$$46 = k_1 (3)^2 + k_2 \frac{1}{2}$$ $$46 = 9k_1 + \frac{1}{2}k_2 \quad \text{(Mlinganyo 2)}$$

Kutoka Mlinganyo 1, gawanya kwa 2:

$$48 = 2k_1 + k_2 \implies k_2 = 48 - 2k_1$$

Badilisha $k_2$ kwenye Mlinganyo 2:

$$46 = 9k_1 + \frac{1}{2}(48 - 2k_1)$$ $$46 = 9k_1 + 24 - k_1$$ $$46 - 24 = 8k_1$$ $$22 = 8k_1$$ $$k_1 = \frac{22}{8} = \frac{11}{4}$$

Sasa tafuta $k_2$:

$$k_2 = 48 - 2\left(\frac{11}{4}\right) = 48 - \frac{11}{2} = \frac{96 - 11}{2} = \frac{85}{2}$$

$$B = \frac{11}{4} M^2 + \frac{85}{2N}$$

Usemi wa B kwa kutumia M na N ni $\boxed{B = \frac{11}{4} M^2 + \frac{85}{2N}}$.