Find four consecutive integers such that the sum of the first and third is eight les than twice the sum of the second and fourth

1 answer:

5 0

X, x+1, x+2, x+3 are 4 consecutive integers

2(x + x+2) = 5(x+3) - 26
2(2x+2) = 5(x+3) - 26
4x + 4 = 5x + 15 - 26
4x + 4 = 5x - 11
15 = x
15, 16, 17, 18

x, x+2, x+4 consecutive even integers
2(x+4)+11 = x + 9
2x + 8 + 11 = x + 9
2x + 19 = x + 9
x = -10
-10, -8, -6

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Answer:

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Step-by-step explanation:

$[(p^2) (q^{-3}) ]^{-2}.[(p)^{-3}(q)^5] ^{-2}\\\\=[(p^2) (q^{-3}) \times(p)^{-3}(q)^5 ]^{-2}\\\\=[(p^{2}) \times(p)^{-3} \times (q^{-3}) \times(q)^5 ]^{-2}\\\\=[(p^{2-3}) \times (q^{5-3}) ]^{-2}\\\\=[(p^{-1}) \times (q^{2}) ]^{-2}\\\\=(p^{-1\times (-2)}) \times (q^{2\times (-2) }) \\\\=p^{2}\times q^{-4} \\\\= \frac{p^2}{q^4}\\\\= \frac{(-2)^2}{(-1)^4}\\\\= \frac{4}{1}\\\\= 4$