A. dog and a cat are 200 meters apart when they see each other. the dog can run at a speed of 30 m/sec, while the cat can run at a speed of 24 m/sec. how soon will the dog catch the cat if the dog starts running after the cat?

33 ¹/₃ seconds

Step-by-step explanation:

Since the dog is 30 - 24 = 6 m/s faster than the cat, that means that the dog is 6 meters closer to the cat each second.

So, that gap will close in 200/6 = 33 ¹/₃ seconds.

(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

step-by-step explanation: s