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how much would u exceed the speed limit for the driving time to be 8 hours

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2 Answers

253/(v +70) + 413/(V +75) =
8 [ 253( V+75)+ 413( V+70)] =
8(V +70)( V +75) 253 V + 413V + 253( 75)+ 413(70)= 8( V^2+ 145V +5250) 666v + 47885 =
8 V^2 + 1160V + 42000=
8 V^2 + 494 V - 5885= 0
V = -494/16ñ( 494^2+ 32(5885)/16
V = -30.8 ± 41.09
V = 41.09 - 30.8 = 11.01
8 = 253/(v+70) + 413/(v+75)

Multiply by (v+70)(v+75)

8(v+70)(v+75) = 253(v+75) + 413(v+70)

8(v^2 + 145v + 70*75) = v(253 + 413) + 253(75) + 413(70)

8v^2 + 8*145v + 8*70*75 = 666v + 47885

8v^2 + 494v - 5885 = 0

a = 8, b = 494, c = -5885

v = (-b ± √(b^2 - 4ac))/(2a)

v = (-494 ± √(494^2 - 4(8)(-5885)))/(2(8))

v = (-494 ± √(432356))/16

v = -494/16 ± √(432356)/16

v ≈ -30.875 ± 41.09611447570196

v ≈ 10.22111447570196 or -41.09611447570196

We want only positive v so

v ≈ 10.22111447570196 mph over the speed limit.
 
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