A train travels 200 km in the same time that a plane covers 900 km. If the speed of the plane is 20 km per hr less than 5 times the speed of the train, find both speeds. r t d Train x 200 Plane y 900

the speed of the train is 40 km/h

the speed of the plane is 180 km/h

Step-by-step explanation:

Let x represent the speed of the train.

If the speed of the plane is 20 km per hr less than 5 times the speed of the train, it means that the speed of the plane would be (5x - 20) km/h

Time = distance/speed

The train travels 200 km. This means that the time taken to travel 200 km is

200/x

in the same time that a plane covers 900 km. This means that the time taken to travel 900 km is

900/(5x - 20)

Since the time is the same, then

200/x = 900/(5x - 20)

Cross multiplying, it becomes

200(5x - 20) = 900x

1000x - 4000 = 900x

1000x - 900x = 4000

100x = 4000

x = 4000/100

x = 40

The speed of the plane would be

5 × 40 - 20

= 180

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