Teo has a combination of quarters, loonies, and toonies in his pocket. He knows he has 9 coins in total and that he has at least one of each
type of coin.
a) hwo many outcomes end with over 8$

b) How many different combinations of coins could Teo use to pay for an item
that costs between $10 and $12, if he uses fewer toonies than loonies?

343

Step-by-step explanation:

There are quarters(25¢), loonies ($1), and toonies ($2)

And total number of coins is 9

If loonies ($1), and toonies ($2) are one, then quarters(25¢) is 7

If quarters(25¢), and toonies ($2) are one, then loonies ($1)is 7

If loonies ($1), and quarters(25¢) are one, then toonies ($2) is 7

So possible number number of combination for an item that cost between $10 and $12 is

= 1^3 * 1^3 * 7^3

= 343

hope this helps, stay safe :)

(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