A train covers certain distance in two parts. Distance covered in first part is 200% more than the distance covered in second part while speed of train is in the ratio 2 : 1 in first and second part respectively. If average speed of train is 64 km/hr, then find the speed of train in first part? (in kmph)

Part A speed=96*2=192km/h

Step-by-step explanation:

Two parts, parts A and part B

Part A=200% more than B

Let part B=x

Part A=200 more than B

=2x

Speed ratio=2:1

Average speed of train=64km/h

Let Part A speed=2x

Part B speed=x

A:B=2:1

Total ratio =3

Speed in the part A can be calculated thus:

Totatl speed= ratio of speed in part A / total ratio

64=2x/3

192=2x

x=96km/h

Part A speed=96*2=192km/h

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

srry idk

step-by-step explanation: