Simplify the following using laws of indices: (a5 * a3) / a4

$\frac{a^5 X a^3}{a^4}$

• Step 1: Tumia sheria ya kuzidisha ya viashiria ($a^m \times a^n = a^{m+n}$) kwa nambari ya juu. $$a^5 \times a^3 = a^{5+3} = a^8$$
• Step 2: Tumia sheria ya kugawanya ya viashiria ($\frac{a^m}{a^n} = a^{m-n}$) kwa matokeo ya hatua ya 1 na nambari ya chini. $$\frac{a^8}{a^4} = a^{8-4} = a^4$$
• Jibu la Mwisho: $$\boxed{\mathbf{a^4}}$$

• Step 1: Acute angle ni pembe ambayo kipimo chake ni chini ya $90^\circ$.
• Step 2: Obtuse angle ni pembe ambayo kipimo chake ni zaidi ya $90^\circ$ lakini chini ya $180^\circ$.
• Jibu la Mwisho: Tofauti ni kwamba pembe kali (acute angle) ni ndogo kuliko $90^\circ$, wakati pembe butu (obtuse angle) ni kubwa kuliko $90^\circ$ lakini ndogo kuliko $180^\circ$.

$x + y = 9 \quad \text{(Equation 1)}$
$x - y = 3 \quad \text{(Equation 2)}$

• Step 1: Kutoka Equation 2, toa $y$ ili kupata $x$ kwa kutumia $y$: $$x = 3 + y \quad \text{(Equation 3)}$$
• Step 2: Badilisha Equation 3 kwenye Equation 1: $$(3 + y) + y = 9$$
• Step 3: Tatua kwa $y$: $$3 + 2y = 9$$ $$2y = 9 - 3$$ $$2y = 6$$ $$y = \frac{6}{2}$$ $$y = 3$$
• Step 4: Badilisha thamani ya $y = 3$ kwenye Equation 3 ili kupata $x$: $$x = 3 + 3$$ $$x = 6$$
• Jibu la Mwisho: $$\boxed{\mathbf{x=6, y=3}}$$

$\sqrt[3]{64} + \sqrt{81}$

• Step 1: Tafuta mzizi wa tatu wa 64. $$\sqrt[3]{64} = 4 \quad (\text{kwa sababu } 4 \times 4 \times 4 = 64)$$
• Step 2: Tafuta mzizi wa pili wa 81. $$\sqrt{81} = 9 \quad (\text{kwa sababu } 9 \times 9 = 81)$$
• Step 3: Jumlisha matokeo. $$4 + 9 = 13$$
• Jibu la Mwisho: $$\boxed{\mathbf{13}}$$

• Step 1: Tumia fomula ya kubadilisha kutoka Celsius kwenda Kelvin: $K = ^\circ C + 273.15$. $$K = 25 + 273.15$$ $$K = 298.15$$
• Jibu la Mwisho: $$\boxed{\mathbf{298.15 K}}$$

b) Convert $293K$ to Celsius.

• Step 1: Tumia fomula ya kubadilisha kutoka Kelvin kwenda Celsius: $^\circ C = K - 273.15$. $$^\circ C = 293 - 273.15$$ $$^\circ C = 19.85$$
• Jibu la Mwisho: $$\boxed{\mathbf{19.85^\circ C}}$$

• Step 1: Fikiria pembetatu mraba yenye pande $a$ (kinyume), $b$ (karibu), na $c$ (hypotenuse), na pembe $x$.
• Step 2: Kwa mujibu wa ufafanuzi wa sine na cosine: $$\sin x = \frac{\text{kinyume}}{\text{hypotenuse}} = \frac{a}{c}$$ $$\cos x = \frac{\text{karibu}}{\text{hypotenuse}} = \frac{b}{c}$$
• Step 3: Kwa mujibu wa Theorem ya Pythagoras katika pembetatu mraba: $$a^2 + b^2 = c^2$$
• Step 4: Gawanya pande zote mbili za equation ya Pythagoras kwa $c^2$: $$\frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{c^2}{c^2}$$ $$\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1$$
• Step 5: Badilisha $\frac{a}{c}$ na $\sin x$, na $\frac{b}{c}$ na $\cos x$: $$(\sin x)^2 + (\cos x)^2 = 1$$ $$\sin^2 x + \cos^2 x = 1$$
• Jibu la Mwisho: $$\boxed{\mathbf{\sin^2 x + \cos^2 x = 1}}$$

$4x - 7 = 13$

• Step 1: Ongeza 7 pande zote mbili za equation. $$4x - 7 + 7 = 13 + 7$$ $$4x = 20$$
• Step 2: Gawanya pande zote mbili kwa 4. $$\frac{4x}{4} = \frac{20}{4}$$ $$x = 5$$
• Jibu la Mwisho: $$\boxed{\mathbf{x=5}}$$

$\log_{10} 1000$

• Step 1: Andika 1000 kama nguvu ya 10. $$1000 = 10^3$$
• Step 2: Badilisha hii kwenye usemi wa logarithm. $$\log_{10} 10^3$$
• Step 3: Tumia sheria ya logarithm $\log_b b^n = n$. $$\log_{10} 10^3 = 3$$
• Jibu la Mwisho: $$\boxed{\mathbf{3}}$$

• Step 1: Kumbuka ufafanuzi wa $\sinh x$: $$\sinh x = \frac{e^x - e^{-x}}{2}$$
• Step 2: Badilisha $x$ na $-x$ katika ufafanuzi: $$\sinh(-x) = \frac{e^{-x} - e^{-(-x)}}{2} = \frac{e^{-x} - e^x}{2}$$
• Step 3: Linganisha $\sinh(-x)$ na $\sinh x$: $$\sinh(-x) = -\left(\frac{e^x - e^{-x}}{2}\right) = -\sinh x$$
• Jibu la Mwisho: Kwa kuwa $\sinh(-x) = -\sinh x$, kazi ya $\sinh x$ ni kazi isiyo ya kawaida (odd function).

b) A cylinder has a radius of 3m and has a height of 7m. Calculate the surface area.

• Step 1: Fomula ya eneo la uso la silinda ni: $$A = 2\pi r^2 + 2\pi rh$$ ambapo $r$ ni radius na $h$ ni urefu.
• Step 2: Badilisha maadili yaliyopewa ($r = 3 \text{ m}$, $h = 7 \text{ m}$) kwenye fomula. $$A = 2\pi (3 \text{ m})^2 + 2\pi (3 \text{ m})(7 \text{ m})$$ $$A = 2\pi (9 \text{ m}^2) + 2\pi (21 \text{ m}^2)$$ $$A = 18\pi \text{ m}^2 + 42\pi \text{ m}^2$$
• Step 3: Jumlisha maneno. $$A = (18 + 42)\pi \text{ m}^2$$ $$A = 60\pi \text{ m}^2$$
• Jibu la Mwisho: $$\boxed{\mathbf{60\pi \text{ m}^2}}$$

• Step 1: Mzunguko wa pembetatu ni jumla ya urefu wa pande zake. $$\text{Perimeter} = \text{Side 1} + \text{Side 2} + \text{Side 3}$$
• Step 2: Badilisha maadili yaliyopewa kwenye fomula. $$24 \text{ cm} = X + (X+2) + (X+4)$$ $$24 = 3X + 6$$
• Step 3: Tatua kwa $X$. $$24 - 6 = 3X$$ $$18 = 3X$$ $$X = \frac{18}{3}$$ $$X = 6$$
• Step 4: Tafuta urefu wa pande tatu. $$\text{Side 1} = X = 6 \text{ cm}$$ $$\text{Side 2} = X+2 = 6+2 = 8 \text{ cm}$$ $$\text{Side 3} = X+4 = 6+4 = 10 \text{ cm}$$
• Jibu la Mwisho: $$\boxed{\mathbf{Pande ni 6 \text{ cm}, 8 \text{ cm}, \text{na } 10 \text{ cm}}}$$

b) The perimeter of an equilateral triangle is 4 cm. The length of one side is expressed as (2x-1)cm, find x.

• Step 1: Katika pembetatu ya usawa (equilateral triangle), pande zote tatu zina urefu sawa. Kwa hivyo, mzunguko ni mara 3 urefu wa upande mmoja. $$\text{Perimeter} = 3 \times \text{Side}$$
• Step 2: Badilisha maadili yaliyopewa kwenye fomula. $$4 \text{ cm} = 3 \times (2x-1) \text{ cm}$$
• Step 3: Tatua kwa $x$. $$4 = 6x - 3$$ $$4 + 3 = 6x$$ $$7 = 6x$$ $$x = \frac{7}{6}$$
• Jibu la Mwisho: $$\boxed{\mathbf{x=\frac{7}{6}}}$$