This is the previous problem: A rabbit started at point 0 and made 3 jumps a jump of 4 units then a jump of 2 units and then a jump of 1 unit we do not know the direction of either jump where on the line could the rabbit be. This is my question: next time, the rabbit from the previous problem started at 10. Where on the number line can he be after the same three jumps?

Q1: The rabbit could be at eight different numbers of the number line: -7, - 5, - 3, - 1, 1, 3, 5, and 7.

Q2: 13

Step-by-step explanation:

Q1: Let's simulate each of the jumps of the rabbit in all the possible directions, as follows:

Option 1: - 4 - 2 - 1 = -7

Option 2: - 4 - 2 + 1 = -5

Option 3: - 4 + 2 - 1 = - 3

Option 4: - 4 + 2 + 1 = - 1

Option 5:  4 - 2 - 1 = 1

Option 6: 4 - 2 + 1 = 3

Option 7: 4 + 2 - 1 = 5

Option 8: 4 + 2 + 1 = 7

Q2: If the rabbit started at 10 you would add 3 to 10 and get 13.

(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

there would only be 50 odd numbers between 1 and 100 so therefor there would be 50 even so that would only be 50 odd numbers.