Hello,
Very nice as problem.
2 solutions:
1 quater,8 dimes, 2 pennies
and
3 quaters,3 dimes, 2 pennies
since
107=( 0, 0, 107) but : 100= 0*25+ 0*10+ 100
107=( 0, 1, 97) but : 100= 0*25+ 1*10+ 90
107=( 0, 2, 87) but : 100= 0*25+ 2*10+ 80
107=( 0, 3, 77) but : 100= 0*25+ 3*10+ 70
107=( 0, 4, 67) but : 100= 0*25+ 4*10+ 60
107=( 0, 5, 57) but : 100= 0*25+ 5*10+ 50
107=( 0, 6, 47) but : 100= 0*25+ 6*10+ 40
107=( 0, 7, 37) but : 100= 0*25+ 7*10+ 30
107=( 0, 8, 27) but : 100= 0*25+ 8*10+ 20
107=( 0, 9, 17) but : 100= 0*25+ 9*10+ 10
107=( 0, 10, 7) but : 100= 0*25+ 10*10+ 0
107=( 1, 0, 82) but : 100= 1*25+ 0*10+ 75
107=( 1, 1, 72) but : 100= 1*25+ 1*10+ 65
107=( 1, 2, 62) but : 100= 1*25+ 2*10+ 55
107=( 1, 3, 52) but : 100= 1*25+ 3*10+ 45
107=( 1, 4, 42) but : 100= 1*25+ 4*10+ 35
107=( 1, 5, 32) but : 100= 1*25+ 5*10+ 25
107=( 1, 6, 22) but : 100= 1*25+ 6*10+ 15
107=( 1, 7, 12) but : 100= 1*25+ 7*10+ 5
107=( 1, 8, 2) is good
107=( 2, 0, 57) but : 100= 2*25+ 0*10+ 50
107=( 2, 1, 47) but : 100= 2*25+ 1*10+ 40
107=( 2, 2, 37) but : 100= 2*25+ 2*10+ 30
107=( 2, 3, 27) but : 100= 2*25+ 3*10+ 20
107=( 2, 4, 17) but : 100= 2*25+ 4*10+ 10
107=( 2, 5, 7) but : 100= 2*25+ 5*10+ 0
107=( 3, 0, 32) but : 100= 3*25+ 0*10+ 25
107=( 3, 1, 22) but : 100= 3*25+ 1*10+ 15
107=( 3, 2, 12) but : 100= 3*25+ 2*10+ 5
107=( 3, 3, 2) is good
107=( 4, 0, 7) but : 100= 4*25+ 0*10+ 0
3 + [_] ÷ 7 = 9
[_] ÷ 7 = 9 - 3
[_] = 6 x 7
[_] = 42
to check, just put 42 in the empty place and solve
Answer:
Range is number of copies produced and set of values is; 1 ≤ N ≤ 200
Domain; Cost of publishing book in dollars; set of values are; $710 ≤ N ≤ $2700
Step-by-step explanation:
Range is a set of all the possible output values in a function while domain is the set of all possible input values.
Now, the function is given as;
C = 10N + 700
Where;
C is the cost of publishing the book in dollars
N is the number of copies of books produced
Thus, the domain will be a set of N values while Range will be a set of C values.
We are told that the first printing can produce up to 200 copies of the book.
That means a maximum of 200 books and a minimum of 1.
Thus;
Range is; 1 ≤ N ≤ 200
Maximum possible cost of the 200 books is;
C = 10(200) + 700
C = $2700
Minimum cost which will be for 1 book will be;
C = 10(1) + 700
C = $710
Thus,domain is;
$710 ≤ N ≤ $2700