Comprehensive of the Sankes
1. Write down three ways in Paragraph 1 in which the writer tells you that Aloo felt
alarmed and frightened as she approached her hut.
2. Explain who Anyango, Ouma, and Akoth are. (Paragraph 2)
3. Why was Aloo so sure that some misfortune had befallen the children? (Paragraph 2)
4. What made aloo certain that one of her children had 'met with fate'? (line 20
5. Explain why Aloo whispered the words, 'What has happened to my baby?', rather
than saying them in a louder voice. (line 23)
Answer need urgently correct answer will make as brainlist
and I will report wrong answers.
Answers
Answer:
So, as a 3d = 10 --1
So, as a 3d = 10 --1Substitute n = 11
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)d
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10d
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth term
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1a10d 31 --2
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1a10d 31 --2Subtract 1 form 2
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1a10d 31 --2Subtract 1 form 2a10da4d 31 10
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1a10d 31 --2Subtract 1 form 2a10da4d 31 106d=21
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1a10d 31 --2Subtract 1 form 2a10da4d 31 106d=21So, d = 21/6 =3.5
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1a10d 31 --2Subtract 1 form 2a10da4d 31 106d=21So, d = 21/6 =3.5Substitute value of d in 1
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1a10d 31 --2Subtract 1 form 2a10da4d 31 106d=21So, d = 21/6 =3.5Substitute value of d in 1a + 3(3.5) = 10
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1a10d 31 --2Subtract 1 form 2a10da4d 31 106d=21So, d = 21/6 =3.5Substitute value of d in 1a + 3(3.5) = 10a 10.5 10a = -0.5
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1a10d 31 --2Subtract 1 form 2a10da4d 31 106d=21So, d = 21/6 =3.5Substitute value of d in 1a + 3(3.5) = 10a 10.5 10a = -0.5Formula of sum of first n terms = Sn (2a + (n − 1)d) =
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1a10d 31 --2Subtract 1 form 2a10da4d 31 106d=21So, d = 21/6 =3.5Substitute value of d in 1a + 3(3.5) = 10a 10.5 10a = -0.5Formula of sum of first n terms = Sn (2a + (n − 1)d) =Substitute n = 25
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1a10d 31 --2Subtract 1 form 2a10da4d 31 106d=21So, d = 21/6 =3.5Substitute value of d in 1a + 3(3.5) = 10a 10.5 10a = -0.5Formula of sum of first n terms = Sn (2a + (n − 1)d) =Substitute n = 25So, S25 = 25 (2(-0.5) + (25 − 1)3.5)
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1a10d 31 --2Subtract 1 form 2a10da4d 31 106d=21So, d = 21/6 =3.5Substitute value of d in 1a + 3(3.5) = 10a 10.5 10a = -0.5Formula of sum of first n terms = Sn (2a + (n − 1)d) =Substitute n = 25So, S25 = 25 (2(-0.5) + (25 − 1)3.5)S25 = 1037.5
So, as a 3d = 10 --1Substitute n = 11a11 = a + (11-1)da11 a 10dNow we are given that 11th term is one more three times of the fourth termSo, a11 - 3a4 + 1So, a 10d=3(10) +1a10d 31 --2Subtract 1 form 2a10da4d 31 106d=21So, d = 21/6 =3.5Substitute value of d in 1a + 3(3.5) = 10a 10.5 10a = -0.5Formula of sum of first n terms = Sn (2a + (n − 1)d) =Substitute n = 25So, S25 = 25 (2(-0.5) + (25 − 1)3.5)S25 = 1037.5thus the sum of 25 terms is 1037.5
Answer:
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hope it helps you