Megan has A quarters and B dimes with a total value of $1.95, where A and B are both counting numbers. How many different values of A can Megan have?

The number of quarters can be 1, 3, 5 or 7

Step-by-step explanation:

Megan has A quarters and B dimes.

The value of a quarter is $0.25 and the value of a dime is $0.10

Then we have that:

A*$0.25 + B*$0.10 = $1.95

We want to know the different possible values of A.

Now, it is usefull to see the different multilpes of 0.25 (this is the odd multyples of 0.25) we only care for the ones that have a 5 in the undredth place, becauses we only can ad multiple of $0.10, so the 5 in the undredth place needs to come from this

Then the possible values of A are the odd numbers such that A*0.25 is smaller than 1.95, let's do the math:

0.25*1 = 0.25 (here 1.95 - 0.25 = 1.70, then B = 1.70/0.10 = 17) the pair is B = 17 and A = 1

0.25*3 = 0.75 (here 1.95 - 75 = 1.20, then B = 1.20/0.10 = 12) the pair is B = 12 and A = 3.

0.25*5 = 1.25 (here 1.95 - 1.25 = 0.70, then B = 0.70/0.10 = 7) the pair is B = 7 and A = 5

0.25*7 = 1.75 (here 1.95 - 1.75 = 0.20, then b = 0.20/0.10 = 2) The pair is B = 2 and A = 7

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)