The shortest route from London to Oxford is 55 miles. A lorry is expected to take 1.1 hours to travel this route.
The lorry actually travels by a different route which increases the distance by 15%, but it still arrives in 1.1 hours.
By how many more mph than the expected speed does the lorry travel?

so to find the mph of the lorry for the original route we divide 66 by 55 since it 66 is 1.1 of 60

66 divided by 55=1.2

so it takes 1 minutes 12 seconds for the lorry to go a mile

now we multiply 55 by 1.15=63.25

so we divide 66 by 63.25=1.04347826087

so it takes 1 minute and 1 second for the the lorry to go a miles

1 minute 1 second is 59 miles per hour

1 minute 12 seconds is 50 miles per hour

so the lorry travels 9 mph over its expected speed

Hope This Helps!!!

Explanation:

distance = rate*time

d = r*t

r = d/t

r = 55/1.1

r = 50

The lorry's original speed is 50 mph when going the original route.

Now consider the longer route, which is 15% longer compared to the original 55 mile route. So the longer route is 1.15*55 = 63.25 miles exactly. Or you could say 15% of 55 = 0.15*55 = 8.25 which adds onto the original 55 to get 55+8.25 = 53.25; either way the longer distance is 63.25 miles.

Computing the new rate or speed gets us

r = d/t

r = 63.25/1.1

r = 57.5

When traveling the original route, the lorry goes 50 mph. When traveling the longer route, the lorry goes 57.5 mph. This is a difference of 57.5 - 50 = 7.5 mph

Meaning that the lorry must drive 7.5 mph faster on the longer route compared to the shortest route. This is if the driver wants to make the trip in the same 1.1 hour timeframe.

Note: 1.1 hours = 1.1*60 = 66 minutes = 1 hour, 6 minutes.

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

wedfghjkl;

tyuiop[]ghjkl;

step-by-step explanatio89