Ajet plane travels 2 times the speed of a commercial airplane. the distance between vancouver
and regina is 1730 km. if the flight from vancouver to regina on a commercial airplane takes
140 minutes longer than a jet plane, what is the time of a commercial plane ride of this route? ( show steps: ))

The time of a commercial airplane is 280 minutes

Step-by-step explanation:

Let

x > the speed of a commercial airplane

y > the speed of a jet plane

t > the time that a jet airplane takes  from Vancouver to Regina

we know that

The speed is equal to divide the distance by the time

y=2x > equation A

The speed of a commercial airplane is equal to

x=1,730/(t+140) > equation B

The speed of a jet airplane is equal to

y=1,730/t > equation C

substitute equation B and equation C in equation A

1,730/t=2(1,730/(t+140))

Solve for t

1/t=(2/(t+140))

t+140=2t

2t-t=140

t=140 minutes

therefore

The time of a commercial airplane is

t+140=140+140=280 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 !

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!