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zysi [14]
3 years ago
15

Can anybody help me in this​

Mathematics
1 answer:
Kobotan [32]3 years ago
8 0

Answer:

a) I donot agree with him because as we see the sequence nicely we will find an interval of 8 between any two terms and 32 and 33 don't match them

Step-by-step explanation:

b) and c) Solution:

Given,

First term (a) = 4

Common difference (d) = 12-4 = 8

Now,

Since, they are in A.P.

nth term = a + (n-1)d

              = 4 + 8(n-1)

              = 4 + 8n-8

              = 8n-4

Again,

1st method for c)

100th term = a + 99d = 4 + 99*8 = 4 + 792 = 796

200th term = a + 199d = 4 + 199*8 = 4 + 1592 = 1596

2nd method for c)

100th term = 8*100 - 4 = 800 - 4 = 796

200th term = 8*200 - 4 = 1600 - 4 = 1596

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Answer: 38 in.^{2}


Step-by-step explanation:

1/ Find the area of the triangular sides. (1/2 x 2 x 6 = 6 in.^{2} for each triangular side)

2/ Find the area of each of the rectangular sides. In this case, the bottom (base) side is 2x6 = 12 in^{2}. The rectangular side on the left is 2x3 = 6 in.^{2}, and the rectangular side on the right is 4x2 = 8 in.^{2}.

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4 0
4 years ago
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igor_vitrenko [27]

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