serenityarts123

18.08.2020 • 

Mathematics

2. A woman travelled from Lagos to Ondo a distance of 350km. Her average speed on the outward journey was ykm. On the return journey, her average speed was 10km/hr faster and therefore she completed the return journey 37.5minutes earlier. Find A. The value of y B. Get average speed on the return journey in km/hr C. The total time taken for the whole journey in hrs. Good luck

Ответ:

gianni0

31.01.2022

Mathematics

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70 km/h

9 hr 22.5 min

Step-by-step explanation:

Given

Distance = 350 kmSpeed = yReturn speed = y + 10Time difference on return = 37.5 min = 37.5/60 hr = 5/8 hr

Difference in time as equation:

350/y - 5/8 = 350/(y+10)(8*350 -5y)/8y = 350/(y+10)(y+10)(560 - y)= 8y*70560y - y² +5600 - 10y = 560yy² + 10y - 5600=0y = (-10 ± √(100 +4*5600))/2y = (-10 ±150)/2y= 70 y = -80 not considered as negative value

Average speed was 70 km/h

Total time = 350/70 + 350/(70+10) = 5 + 4.375 = 9.375 hr = 9 hr 22.5 min

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Ответ:

kyliearnell

01.04.2020

Mathematics

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-4<-1, and also -1<4

Step-by-step explanation:

the first box and the last box

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