Please help ty! I added extra points.
Hillary and Charlene both drove from City A to City B. At 10 a. m., Hillary left City A and drove at an average speed of 120 km/h. Charlene drove at an average speed of 144 km/h and took 50 minutes. She arrived at City B at the same time as Hillary. Find the time Charlene left City A.

Solution:

Given that,

At 10 a.m., Hillary left City A and drove at an average speed of 120 km/h

Charlene drove at an average speed of 144 km/h and took 50 minutes.

She arrived at City B at the same time as Hillary

Their distance is same

Find the ratio of speed

120 : 144

Reduce to lowest term

Hillary : charlne = 5 : 6

Find the ratio of time:

Since, speed and time are inversely propotional

Hillary : charlne = 6 : 5

Let time taken by Hiliary be 6t

Let time taken by charlene be 5t

Given that, Charlene took 50 minutes

Therefore,

5t = 50

t = 10

Thus,

time taken by Hiliary = 6t = 6(10) = 60

Thus hiliary took 60 minutes to reach

Find the time Charlene left City A

At 10 a.m., Hillary left City A

Difference between both their time = 60 - 50 = 10 minutes

Thus, Hilairy left city A = 10 am + 10 minutes = 10 : 10 a.m

Thus, Hillary left city A at 10: 10 AM

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: