Two points move along a circle of length 120m with a constant speed. if they move in different directions, then they meet every 15 seconds. when going in the same direction, one point catches up to the second every 60 seconds. find the speeds of the points.

Let x and y be the speeds of the two points. The circle is of length 120m that is Circumference of circle is 120m.When the points are moving in opposite directions, they meet in every 15 seconds.That is the points cover 120m in 15 seconds.As they are moving in opposite directions, so the relative speed will be equal to the sum of speeds of both points.Speed= Distance ÷Time = 120÷15=8m per seconds.

x+y=8

When going in the same direction, one point catches up to the second every 60 seconds.Speed= 120÷60=2 m per seconds

x-y=2.

Solving the two equations by elimination method:

x+y=8.

x-y=2

Adding the equations:

2x=10 ,x=5m / sec.

x+y=8

5+y=8

y=3m/sec.

Thus, the speed of the two points is 5 m/s and 3 m/s.

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2