Find the 91st term of the arithmetic sequence 28, 8, -12, ,8,−

Since there are only three terms and they repeat we can easily figure this out.

To do that all we have to do is determine how far away 91 is from a multiple of three. If it lands on a multiple of three it will be -12. If it is two above a multiple of three it will be eight. If it is one above a multiple of three it will be 28.

So the 91st term in one above a multiple of three the answer will be 28.

i think this could be a possible answer (1 / 84) -1

good luck! : d

(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