Apilot flew a jet from city a to city b, a distance of 2300 mi. on the return trip, the average speed was 20% faster than the outbound speed. the round-trip took 8 h 20 min. what was the speed from city a to city b?

The speed from City A to City B is 506 miles per hour

Solution:

A pilot flew a jet from City A to City B, a distance of 2300 miles

Distance for trip = 2300 miles

Distance for return trip = 2300 miles

The round-trip took 8 h 20 min

Total time taken for both trips = 8 hour 20 minutes

We know that,

1 hour = 60 minutes

Thus, we get,

Total time taken for both trips =

On the return trip, the average speed was 20% faster than the outbound speed

Let "x" be the speed for out bound trip

Then, x + 20 % of x is the average speed of return trip

Speed for return trip = 1.2x

Time taken is given by formula:

Thus speed from City A to City B is 506 miles per hour

dilation scale factor 2: starting with δabc, draw the dilation image of the triangle with a center at the origin and a scale factor of two. notice that every coordinate of the original triangle has been multiplied by the scale factor (x2). dilation with scale factor 2, multiply by 2.

shell's the art director at her local community center. she paid $10 and for each child that attends she earns $4.

for each child she ears $4 that is the slope m = 4

she paid $10, that is the initial amount paid. so 10 is the y intercept b

m =4   and b = 10

y intercept is 10 so the point is (0,10)

we start graph from y intercept (0,10)

slope = 4 . we know slope = rise / run

slope = 4   or 4/1   are same

rise = 4   and run = 1

for rise we move 4 units up and for run we move 1 unit to the right

the graph is attached below

x axis is the number of child she attends

y axis is the amount   she earns