It takes the high-speed train x hours to travel the z miles from town a to town b at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. if the high-speed train leaves town a for town b at the same time that the regular train leaves town b for town a, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other? (a) z(y – x)/x + y(b) z(x – y)/x + y(c) z(x + y)/y – x(d) xy(x – y)/x + y(e) xy(y – x)/x + y

S=  z (y-x)/(x+y)

Step-by-step explanation:

Lets the speed of high speed train = u

Lets the regular speed train = v

We know that

Distance = Speed x time

For high speed train

 z = u .x      

    u= z/x                  1

For regular speed train

 z = v .y    

v = z/y             2

Both are traveling in opposite direction so relative speed

Vr = z / x+ z /y

Lets in time t they will meet

z = (z / x+ z /y) t

t= xy/ (x+y)

Lets take distance  cover by high speed train is m when it moves A to B and  speed cover by regular train is n when it is moving B to A.They meet at time t.

m = u .t

m = z / x .xy/ (x+y)

m = zy/ (x+y)            3

n = v .t

n = z / y .xy/ (x+y)

n = zx/ (x+y)              4

From equation 3 and 4

So

m - n= zy/ (x+y)   - zx/ (x+y)      

S=  z (y-x)/(x+y)

Option a is correct.

d

step-by-step explanation:

i used the pitagoras theoreme

let y equal to

the total fare and x is the mile of taxi ride. so the equation is

y = 0.55x + 1.75

since susie has

$10 to spend for a taxi cab, so he can have

10 = 0.55x +

1.75

x = 15 miles of

taxi ride

so the system of

inequality is

10 < 0.55x +

1.75

x > 2