Two lines have equations: L1: 3x – 4y + 12 = 0 and L2: 6x + 8y – 5 = 0. a) Express each line in the form y = mx + c. b) State whether the two lines are parallel, perpendicular or neither, giving a reason.

Toa $3x$ na 12 kutoka pande zote mbili za mlinganyo.

Gawanya kila neno kwa $-4$.

Mlinganyo wa L1 katika fomu $y = mx + c$ ni $\boxed{y = \frac{3}{4}x + 3}$.

Kwa L2: $6x + 8y - 5 = 0$

Toa $6x$ na ongeza 5 kwenye pande zote mbili za mlinganyo.

Gawanya kila neno kwa 8.

Mlinganyo wa L2 katika fomu $y = mx + c$ ni $\boxed{y = -\frac{3}{4}x + \frac{5}{8}}$.

b) State whether the two lines are parallel, perpendicular or neither, giving a reason.

Kutoka sehemu a), mteremko wa L1 ni $m_1 = \frac{3}{4}$.
Mteremko wa L2 ni $m_2 = -\frac{3}{4}$.

Mistari ni sambamba (parallel) ikiwa miteremko yao ni sawa ($m_1 = m_2$). Hapa, $m_1 = \frac{3}{4}$ na $m_2 = -\frac{3}{4}$, kwa hivyo $m_1 \neq m_2$. Mistari sio sambamba.
Mistari ni perpendicular ikiwa bidhaa ya miteremko yao ni $-1$ ($m_1 \cdot m_2 = -1$).

Kwa kuwa $-\frac{9}{16} \neq -1$, mistari sio perpendicular.

Kwa kuwa miteremko sio sawa na bidhaa ya miteremko sio $-1$, mistari sio sambamba wala perpendicular.
Mistari ni $\boxed{\text{neither parallel nor perpendicular}}$ kwa sababu miteremko yao ($m_1 = \frac{3}{4}$ na $m_2 = -\frac{3}{4}$) sio sawa na bidhaa ya miteremko yao ($-\frac{9}{16}$) sio $-1$.

7. The first term of an arithmetic progression (A.P.) is 7 and the fifth term is 19.

a) Find the common difference.

Formula ya neno la $n$-th ni $a_n = a + (n-1)d$, ambapo $a$ ni neno la kwanza na $d$ ni tofauti ya kawaida.

Neno la kwanza, $a = 7$.
Neno la tano, $a_5 = 19$.

Toa 7 kutoka pande zote mbili:

Gawanya kwa 4:

Tofauti ya kawaida ni $\boxed{3}$.

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