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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Huu hapa ni utatuzi wa maswali yako: 7. Pata kinyume cha -2 & 6 \\ 3 & -7 . Kisha tatua milinganyo ifuatayo kwa kutumia njia ya matriki. 3x - 7y - 26 = 0 6y - 2x + 20 = 0 Step 1: Andika upya milinganyo katika mpangilio unaolingana na matriki iliyotolewa na uweke katika fomu ya matriki. Milinganyo iliyopangwa upya: -2x + 6y = -20 3x - 7y = 26 Katika fomu ya matriki AX = B: A = -2 & 6 \\ 3 & -7 , X = x \\ y , B = -20 \\ 26 Step 2: Pata kinyume cha matriki A. Kwanza, kokotoa kiamua (determinant) cha A: (A) = (-2)(-7) - (6)(3) = 14 - 18 = -4 Kinyume cha A ni: A^-1 = (1)/((A)) -7 & -6 \\ -3 & -2 = (1)/(-4) -7 & -6 \\ -3 & -2 A^-1 = (-7)/(-4) & (-6)/(-4) \\ (-3)/(-4) & (-2)/(-4) = (7)/(4) & (3)/(2) \\ (3)/(4) & (1)/(2) Kinyume cha matriki ni (7)/(4) & (3)/(2) \\ (3)/(4) & (1)/(2) . Step 3: Tumia kinyume cha matriki kutatua milinganyo. X = A^-1 B x \\ y = (7)/(4) & (3)/(2) \\ (3)/(4) & (1)/(2) -20 \\ 26 x = ((7)/(4))(-20) + ((3)/(2))(26) = 7(-5) + 3(13) = -35 + 39 = 4 y = ((3)/(4))(-20) + ((1)/(2))(26) = 3(-5) + 1(13) = -15 + 13 = -2 Hivyo, suluhisho ni x=4, y=-2. 8. Mchoro hapa chini unaonyesha duara MNPQR, kituo O. MQ na NR ni mistari iliyonyooka. Ikiwa pembe ROQ = 62°, pata ukubwa wa pembe NPQ. Step 1: Tambua pembe za katikati na uhusiano wao. MQ na NR ni mistari iliyonyooka inayopita katikati O, kwa hivyo ni kipenyo. Pembe ROQ = 62^. Kwa kuwa MQ ni kipenyo (mstari ulionyooka), pembe NOQ na ROQ ni pembe za ziada kwenye mstari NR. No, this is incorrect. MQ and NR are diameters. NOQ na MOR ni pembe zinazopingana kwa wima. NOM na ROQ ni pembe zinazopingana kwa wima. Hivyo, NOM = ROQ = 62^. Pembe NOQ na NOM ni pembe za ziada kwenye kipenyo MQ. NOQ = 180^ - NOM = 180^ - 62^ = 118^. Step 2: Pata pembe NRQ. Pembe NRQ ni pembe kwenye mduara inayotegemea upinde NQ. Pembe ya katikati inayotegemea upinde NQ ni NOQ. Pembe kwenye mduara ni nusu ya pembe ya katikati inayotegemea upinde uleule. NRQ = (1)/(2) NOQ = (1)/(2) × 118^ = 59^ Step 3: Tumia sifa za mstatili wa mduara (cyclic quadrilateral) kupata NPQ. NPQR ni mstatili wa mduara (cyclic quadrilateral). Katika mstatili wa mduara, jumla ya pembe zinazopingana ni 180^. NPQ + NRQ = 180^ NPQ + 59^ = 180^ NPQ = 180^ - 59^ = 121^ Ukubwa wa pembe NPQ ni 121^. 9. Bainisha tofauti (variance) kwa seti ifuatayo ya namba: 6, 10, 7, 6, 9, 3, 2, 9, 8, 5, 4, 4, 5. Step 1: Panga namba na uhesabu idadi ya namba (n). Namba: 2, 3, 4, 4, 5, 5, 6, 6, 7, 8, 9, 9, 10. Idadi ya namba n = 13. Step 2: Kokotoa wastani (x). Jumla ya namba ( x): 2+3+4+4+5+5+6+6+7+8+9+9+10 = 78. Wastani x = ( x)/(n) = (78)/(13) = 6. Step 3: Kokotoa mkengeuko wa mraba kutoka kwa wastani ((x - x)^2) kwa kila namba na jumla yake. | x | x - x | (x - x)^2 | |-----|---------------|-------------------| | 2 | 2-6 = -4 | 16 | | 3 | 3-6 = -3 | 9 | | 4 | 4-6 = -2 | 4 | | 4 | 4-6 = -2 | 4 | | 5 | 5-6 = -1 | 1 | | 5 | 5-6 = -1 | 1 | | 6 | 6-6 = 0 | 0 | | 6 | 6-6 = 0 | 0 | | 7 | 7-6 = 1 | 1 | | 8 | 8-6 = 2 | 4 | | 9 | 9-6 = 3 | 9 | | 9 | 9-6 = 3 | 9 | | 10 | 10-6 = 4 | 16 | Jumla ya (x - x)^2 = 16+9+4+4+1+1+0+0+1+4+9+9+16 = 74. Step 4: Kokotoa tofauti (variance). Fomula ya tofauti ni ^2 = (x - x)^2n. ^2 = (74)/(13) Tofauti ni (74)/(13) (au takriban 5.692). Drop the next question.