This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Haya Dennis, twende kazi! 1. Simplify the following using laws of indices: (a^5 X a^3)/(a^4) Step 1: Tumia sheria ya kuzidisha ya viashiria (a^m × a^n = a^m+n) kwa nambari ya juu. a^5 × a^3 = a^5+3 = a^8 Step 2: Tumia sheria ya kugawanya ya viashiria ((a^m)/(a^n) = a^m-n) kwa matokeo ya hatua ya 1 na nambari ya chini. (a^8)/(a^4) = a^8-4 = a^4 Jibu la Mwisho: a^4 2. State the difference between an acute angle and an obtuse angle. Step 1: Acute angle ni pembe ambayo kipimo chake ni chini ya 90^*. Step 2: Obtuse angle ni pembe ambayo kipimo chake ni zaidi ya 90^ lakini chini ya 180^*. Jibu la Mwisho: Tofauti ni kwamba pembe kali (acute angle) ni ndogo kuliko 90^, wakati pembe butu (obtuse angle) ni kubwa kuliko 90^ lakini ndogo kuliko 180^. 3. Solve the simultaneous equations using substitution method: x + y = 9 (Equation 1) x - y = 3 (Equation 2) Step 1: Kutoka Equation 2, toa y ili kupata x kwa kutumia y: x = 3 + y (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 = (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: x=6, y=3 4. Simplify: [3]64 + sqrt(81) Step 1: Tafuta mzizi wa tatu wa 64. [3]64 = 4 (kwa sababu 4 × 4 × 4 = 64) Step 2: Tafuta mzizi wa pili wa 81. sqrt(81) = 9 (kwa sababu 9 × 9 = 81) Step 3: Jumlisha matokeo. 4 + 9 = 13 Jibu la Mwisho: 13 5. a) Convert 25^ C to Kelvin. Step 1: Tumia fomula ya kubadilisha kutoka Celsius kwenda Kelvin: K = ^ C + 273.15. K = 25 + 273.15 K = 298.15 Jibu la Mwisho: 298.15 K b) Convert 293K to Celsius. Step 1: Tumia fomula ya kubadilisha kutoka Kelvin kwenda Celsius: ^ C = K - 273.15. ^ C = 293 - 273.15 ^ C = 19.85 Jibu la Mwisho: 19.85^ C 6. Starting from a right-angled triangle, prove that ^2 x + ^2 x = 1. 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: x = kinyumehypotenuse = (a)/(c) x = karibuhypotenuse = (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: (a^2)/(c^2) + (b^2)/(c^2) = (c^2)/(c^2) ((a)/(c))^2 + ((b)/(c))^2 = 1 Step 5: Badilisha (a)/(c) na x, na (b)/(c) na x: ( x)^2 + ( x)^2 = 1 ^2 x + ^2 x = 1 Jibu la Mwisho: ^2 x + ^2 x = 1 7. Solve the equation: 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. (4x)/(4) = (20)/(4) x = 5 Jibu la Mwisho: x=5 8. Evaluate using logarithms: _10 1000 Step 1: Andika 1000 kama nguvu ya 10. 1000 = 10^3 Step 2: Badilisha hii kwenye usemi wa logarithm. _10 10^3 Step 3: Tumia sheria ya logarithm _b b^n = n. _10 10^3 = 3 Jibu la Mwisho: 3 9. a) Determine whether x is an odd or even function. Step 1: Kumbuka ufafanuzi wa x: x = e^x - e^-x2 Step 2: Badilisha x na -x katika ufafanuzi: (-x) = e^-x - e^-(-x)2 = e^-x - e^x2 Step 3: Linganisha (-x) na x: (-x) = -(e^x - e^-x2) = - x Jibu la Mwisho: Kwa kuwa (-x) = - x, kazi ya 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 r^2 + 2 rh ambapo r ni radius na h ni urefu. Step 2: Badilisha maadili yaliyopewa (r = 3 m, h = 7 m) kwenye fomula. A = 2 (3 m)^2 + 2 (3 m)(7 m) A = 2 (9 m^2) + 2 (21 m^2) A = 18 m^2 + 42 m^2 Step 3: Jumlisha maneno. A = (18 + 42) m^2 A = 60 m^2 Jibu la Mwisho: 60 m^2 10. a) The sides of a scalene triangle are X, X+2 and X+4. If the perimeter is 24cm, Find the length of 3 sides. Step 1: Mzunguko wa pembetatu ni jumla ya urefu wa pande zake. Perimeter = Side 1 + Side 2 + Side 3 Step 2: Badilisha maadili yaliyopewa kwenye fomula. 24 cm = X + (X+2) + (X+4) 24 = 3X + 6 Step 3: Tatua kwa X. 24 - 6 = 3X 18 = 3X X = (18)/(3) X = 6 Step 4: Tafuta urefu wa pande tatu. Side 1 = X = 6 cm Side 2 = X+2 = 6+2 = 8 cm Side 3 = X+4 = 6+4 = 10 cm Jibu la Mwisho: Pande ni 6 cm, 8 cm, na 10 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. Perimeter = 3 × Side Step 2: Badilisha maadili yaliyopewa kwenye fomula. 4 cm = 3 × (2x-1) cm Step 3: Tatua kwa x. 4 = 6x - 3 4 + 3 = 6x 7 = 6x x = (7)/(6) Jibu la Mwisho: x=(7)/(6) Last free one today — make it count tomorrow, or type /upgrade for unlimited.