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vesna_86 [32]
2 years ago
7

Factorise the following.​

Mathematics
1 answer:
zepelin [54]2 years ago
8 0

Answer:

Step-by-step explanation:

i). \frac{48a^3}{6a}=\frac{6\times 8\times a\times a^2}{6a}

          =\frac{6a}{6a}\times 8a^2

          = 8a²

ii). \frac{72a^3b^4c^5}{8ab^2c^3} = \frac{8\times 9\times a\times a^2\times b^2\times b^2\times c^2\times c^3}{8ab^2c^3}

                = \frac{8ab^2c^3}{8ab^2c^3}\times 9a^2b^2c^2

                = 9a²b²c²

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Answer:

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Case 1:

Distance = 288 km

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Case 2:

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Since 288/x > 288/x + 4

288/x - 288/x+4 = 1

288[1/x - 1/x+4 ] = 1

[ x + 4 - x / x(x + 4) ] = 1/288

[4 / x^2 + 4x ] = 1/288

x^2 + 4x = 1152

x^2 + 4x - 1152 = 0

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(x + 36)(x - 32) = 0

x + 36 = 0 , x - 32 = 0

x = -36 , x = 32

x = -36 , rejected since speed cannot be negative.

Therefore , speed of the train = 32 km/h

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