Given that 16a2 + ba + 9 is a perfect square, simplify (16a2+ba+9)/(16a2-9).

Usemi $16a^2 + ba + 9$ ni mraba kamili. Hii inamaanisha unaweza kuandikwa kama $(Ax \pm B)^2$.
Kulinganisha na $A^2x^2 \pm 2ABx + B^2$:
$A^2 = 16 \implies A = 4$.
$B^2 = 9 \implies B = 3$.
Hivyo, $16a^2 + ba + 9 = (4a \pm 3)^2$.
Kiasi cha chini $16a^2-9$ ni tofauti ya miraba, ambayo inaweza kuandikwa kama $(4a)^2 - 3^2 = (4a-3)(4a+3)$.

Kutoka $(4a \pm 3)^2$, kiasi cha kati ni $\pm 2(4a)(3) = \pm 24a$.
Hivyo, $ba = 24a$ au $ba = -24a$. Hii inamaanisha $b = 24$ au $b = -24$.

• Kesi ya 1: Ikiwa $b = 24$ Numerator inakuwa $16a^2+24a+9 = (4a+3)^2$.

  $$\frac{16a^2+24a+9}{16a^2-9} = \frac{(4a+3)^2}{(4a-3)(4a+3)}$$

  Rahisisha kwa kufuta $(4a+3)$ moja kutoka juu na chini:

  $$\frac{4a+3}{4a-3}$$

• Kesi ya 2: Ikiwa $b = -24$ Numerator inakuwa $16a^2-24a+9 = (4a-3)^2$.

  $$\frac{16a^2-24a+9}{16a^2-9} = \frac{(4a-3)^2}{(4a-3)(4a+3)}$$

  Rahisisha kwa kufuta $(4a-3)$ moja kutoka juu na chini:

  $$\frac{4a-3}{4a+3}$$

Kwa hivyo, kuna majibu mawili yanayowezekana kulingana na thamani ya $b$:
Majibu: $\boxed{\frac{4a+3}{4a-3} \text{ au } \frac{4a-3}{4a+3}}$

5. Solve for $x$, given that $\sin 2x = \frac{1}{2}$ for $0 \le x \le 180^\circ$.

Tunapewa $\sin 2x = \frac{1}{2}$.
Thamani kuu ya $2x$ ambapo sine ni $\frac{1}{2}$ ni $30^\circ$.
Kwa kuwa sine ni chanya, $2x$ inaweza kuwa katika roboduara ya kwanza au ya pili.
Hivyo, $2x = 30^\circ$ au $2x = 180^\circ - 30^\circ = 150^\circ$.

Masafa ya $x$ ni $0^\circ \le x \le 180^\circ$.
Hii inamaanisha masafa ya $2x$ ni $0^\circ \le 2x \le 360^\circ$.
Kutoka hatua ya 1, tuna suluhisho mbili za kwanza:
$2x = 30^\circ \implies x = \frac{30^\circ}{2} = 15^\circ$.
$2x = 150^\circ \implies x = \frac{150^\circ}{2} = 75^\circ$.

Tunahitaji pia kuzingatia mizunguko mingine ndani ya masafa ya $2x$.
Kwa $2x = 30^\circ + n \cdot 360^\circ$:
Ikiwa $n=1$, $2x = 30^\circ + 360^\circ = 390^\circ$. Hii inatoa $x = 195^\circ$, ambayo iko nje ya masafa $0^\circ \le x \le 180^\circ$.

Kwa $2x = 150^\circ + n \cdot 360^\circ$:
Ikiwa $n=1$, $2x = 150^\circ + 360^\circ = 510^\circ$. Hii inatoa $x = 255^\circ$, ambayo iko nje ya masafa $0^\circ \le x \le 180^\circ$.

Hivyo, suluhisho pekee ndani ya masafa yaliyotolewa ni $15^\circ$ na $75^\circ$.

Jibu: $\boxed{x = 15^\circ, 75^\circ}$

6. Find the percentage error in the calculation of $4.5\text{cm} \times 5.5\text{cm} + 1.1\text{cm}$.

Thamani halisi ya vipimo ni $4.5\text{cm}$, $5.5\text{cm}$, na $1.1\text{cm}$.
Hesabu ni $P = 4.5\text{cm} \times 5.5\text{cm} + 1.1\text{cm}$.

Kwa kuwa vipimo vimetolewa kwa sehemu moja ya desimali, kosa kamili ni nusu ya kitengo kidogo zaidi, yaani $0.05\text{cm}$.
$\Delta L_1 = \pm 0.05\text{cm}$ kwa $4.5\text{cm}$.
$\Delta L_2 = \pm 0.05\text{cm}$ kwa $5.5\text{cm}$.
$\Delta L_3 = \pm 0.05\text{cm}$ kwa $1.1\text{cm}$.

Hebu $Q = L_1 \times L_2$. Kosa kamili katika bidhaa huhesabiwa kama $\Delta Q = L_1 \Delta L_2 + L_2 \Delta L_1$.

Hivyo, $Q = 24.75 \pm 0.50\text{cm}^2$.

Kwa kujumlisha, makosa kamili huongezwa: $\Delta P = \Delta Q + \Delta L_3$.

Asilimia ya kosa = $\frac{\text{Kosa kamili}}{\text{Thamani halisi}} \times 100\%$.

Kuzungusha hadi sehemu tatu muhimu (3 significant figures):

Jibu: $\boxed{2.13\%}$