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Secondary 1 | Maths
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thanks for the help:)
Date Posted: 5 years ago
Views: 275
Q11: use algebra. Let the distance cycled be x. Hence distance walked will be 150-x km
so, constructing an equation,
x/35+ (150-x)/5= 4.5
6x= 892.5
x= 148.75km
Q12: circumference of wheel = 48 x pi cm = 150.816
speed of car= 3.5km/h = 350000 cm/ h = 5833.3333 cm/min
therefore, number of revolutions = 5833.333/150.816=38.6=39(nearest whole number)
so, constructing an equation,
x/35+ (150-x)/5= 4.5
6x= 892.5
x= 148.75km
Q12: circumference of wheel = 48 x pi cm = 150.816
speed of car= 3.5km/h = 350000 cm/ h = 5833.3333 cm/min
therefore, number of revolutions = 5833.333/150.816=38.6=39(nearest whole number)
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Question 11 :)
Date Posted:
5 years ago
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Question 12 :D
Date Posted:
5 years ago
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