Victor has quarters, dimes and pennies in his pocket. He has three times fewer pennies than dimes and he has 4 fewer quarters than pennies. How many coins does he have total if they are worth $4.60?

Let's define the variables :

Q = number of quarters that Victor has

P = number of pennies that victor has

D = number of dimes that victor has.

We know that:

"He has three times fewer pennies than dimes"

P = D*3

"he has 4 fewer quarters than pennies."

Q = P - 4

And we know that in total, he has $4.60, then:

Q*$0.25 + P*$0.10 + D*0.05 = $4.60

Then we have a system of equations:

P = D/3

Q = P - 4

Q*$0.25 + P*$0.01 + D*0.10 = $4.60

To solve this, we need to replace variables in order to eliminate them.

We can start by replacing the first equation into the other two:

Q = D/3 - 4

Q*$0.25 + (D/3)*$0.01 + D*0.10 = $4.60

Now we can replace the top equation into the bottom one:

(D/3 - 4)*$0.25 + (D/3)*$0.01 + D*0.10 = $4.60

And now solve this for D:

D*($0.25/3 + $0.01/3 + $0.10) -$1  = $4.60

D*$0.1866... = $4.60 + $1.00 = $5.60

D = $5.60/$0.1866... = 30

Then he has 30 dimes.

P = D/3 = 30/3 = 10 pennies

Q = P - 4 = 10 - 4 = 6 quarters.

In total he has 30 + 10 + 6 = 46 coins.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

it's