Alexandra has some dimes and some quarters. She has no more than 21 coins worth a minimum of $3.75 combined. If Alexandra has 4 dimes, determine the minimum number of quarters that she could have. If there are no possible solutions, submit an empty answer.

A minimum of 10 dimes and 11 quarters is what Alexandra will have

Step-by-step explanation:

Let

d = number of dimes

q = number of quarters

Since she has 21 coins altogether,

d + q = 21equation 1    

0.10 x d + 0.25 x q = 3.75

where $.10 is the value of one dime and $.25 is the value of one quarter

make d the subject of formula from equation 1 d = 21 -qequation 3

 insert it in equation 2

0.10d +0.25q =3.75

0.10(21-q) + 0.25q = 3.75

0.1(21)-0.1q+0.25q=3.75

2.1 +0.15q = 3.75

0.15q  = 3.75-2.1 = 1.65

q = 1.65/0.15 =165/15 =11

d = 21 - q

d = 21-11

d = 10

Alexandra has 10 dimes and 11 quarters.

from my calculation i can see that the a minimum of 10 dimes and 11 quarters is what Alexandra will have

the volume is the product of the area of the base, 3x(3x+4), and the height, 3x-1

step-by-step explanation:

125 yards

step-by-step explanation:

i take 75 plus 50.