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 hapa masuluhisho ya maswali yako: Swali la 13: G.C.D. na L.C.M. ya nambari tatu ni 3 na 1008 mtawalia. Ikiwa nambari mbili kati ya hizo ni 48 na 72, tafuta thamani ndogo kabisa inayowezekana ya nambari ya tatu. Step 1: Andika nambari zilizopewa na G.C.D. na L.C.M. katika vizidisho vikuu. Tunazo nambari N_1 = 48, N_2 = 72. Acha nambari ya tatu iwe N_3 = c. G.C.D. = 3 = 2^0 × 3^1 × 7^0 L.C.M. = 1008 = 2^4 × 3^2 × 7^1 Vizidisho vikuu vya nambari zilizopewa: N_1 = 48 = 2^4 × 3^1 × 7^0 N_2 = 72 = 2^3 × 3^2 × 7^0 Acha N_3 = c = 2^x × 3^y × 7^z. Step 2: Tumia sifa za G.C.D. na L.C.M. kwa kila kizidisho kikuu. Kwa G.C.D., exponent ya kila kizidisho kikuu ni ndogo zaidi ya exponents za vizidisho hivyo katika nambari zote. Kwa L.C.M., exponent ya kila kizidisho kikuu ni kubwa zaidi ya exponents za vizidisho hivyo katika nambari zote. Kwa kizidisho kikuu 2: (4, 3, x) = 0 x = 0 (4, 3, x) = 4 (4, 3, 0) = 4. Hii inathibitisha x=0. Kwa kizidisho kikuu 3: (1, 2, y) = 1 y 1 (1, 2, y) = 2 y 2 Hivyo, y inaweza kuwa 1 au 2. Kwa kizidisho kikuu 7: (0, 0, z) = 0 (0, 0, z) = 1 z = 1 Hivyo, z=1. Step 3: Pata thamani zinazowezekana za nambari ya tatu c. Kutoka hatua ya 2, c = 2^x × 3^y × 7^z ambapo x=0, y \1, 2\, na z=1. Ikiwa y=1: c = 2^0 × 3^1 × 7^1 = 1 × 3 × 7 = 21. Ikiwa y=2: c = 2^0 × 3^2 × 7^1 = 1 × 9 × 7 = 63. Step 4: Chagua thamani ndogo kabisa inayowezekana. Thamani ndogo kabisa kati ya 21 na 63 ni 21. Nambari ya tatu ni 21. Swali la 14: Mstari ulionyooka unaopitia pointi A (2, 1) na B (4, m) ni perpendicular kwa mstari ambao equation yake ni 3y = 5-2x. Tafuta thamani ya m. Step 1: Tafuta gradient ya mstari uliotolewa. Equation ya mstari uliotolewa ni 3y = 5-2x. Iandike upya katika fomu y = kx + c: y = (5)/(3) - (2)/(3)x Gradient ya mstari huu, k_1, ni -(2)/(3). Step 2: Tafuta gradient ya mstari unaopitia A na B. Kwa kuwa mistari miwili ni perpendicular, bidhaa ya gradients zao ni -1. Acha gradient ya mstari unaopitia A na B iwe k_2. k_1 × k_2 = -1 -(2)/(3) × k_2 = -1 k_2 = (-1)/(-2)3 = (3)/(2) Step 3: Tumia pointi A (2, 1) na B (4, m) kupata k_2. Fomula ya gradient kati ya pointi mbili (x_1, y_1) na (x_2, y_2) ni k = (y_2 - y_1)/(x_2 - x_1). k_2 = (m - 1)/(4 - 2) = (m - 1)/(2) Step 4: Linganisha semi mbili za k_2 na utatue kwa m. (m - 1)/(2) = (3)/(2) Panua pande zote mbili kwa 2: m - 1 = 3 Ongeza 1 kwa pande zote mbili: m = 3 + 1 m = 4 Thamani ya m ni 4. Swali la 15: Tumia jedwali la reciprocals pekee kukokotoa (3)/(0.6735) + (13)/(0.156) Step 1: Andika upya maneno kama kuzidisha kwa reciprocals. (3)/(0.6735) + (13)/(0.156) = 3 × (1)/(0.6735) + 13 × (1)/(0.156) Step 2: Tafuta reciprocals za nambari zilizopewa (kama vile kutoka jedwali la reciprocals). (1)/(0.6735) ≈ 1.484809 (1)/(0.156) ≈ 6.410256 Step 3: Fanya kuzidisha. 3 × 1.484809 = 4.454427 13 × 6.410256 = 83.333328 Step 4: Jumlisha matokeo. 4.454427 + 83.333328 = 87.787755 Kukokotoa kwa usahihi wa sehemu nne za desimali: 87.7878 Jumla ni 87.7878. Tuma swali linalofuata 📸