<span><em>12 pennies, 3 nickles, and 2 dimes</em>
p = number of pennies
n = number of nickles
d = number of dimes
p(1) + n(5) + d(10) = 47
that is, the number of pennies x 1 cent + number nickles x 5 cents
+ number of dimes x ten cents equals 47 cents
p = 4n
p + n + d = 17
Substituting 4n for p in the above
4n + n + d = 17
5n + d = 17
Subtract 5n from each side
d = 17 - 5n
We will now substitute 4n for p and ( 17-5n ) for d in
the equation
p(1) + n(5) + d(10) = 47
4n(1) +n(5) + (17-5n)(10) = 47
9n + 170 - 50n = 47
-41n + 170 = 47
Subtract 170 from each side
-41n = 47 - 170
-41n = -123
Divide each side by -41
n = 3
Since p = 4n
p = 4(3)
p = 12
Since p + n + d = 17
12 + 3 + d = 17
15 + d = 17
d = 2
So we have 12 pennies, 3 nickles and 2 dimes
12 + 3(5) + 2(10) ?= 47
12 + 15 + 20 ?= 47</span>
The cc’d between 23 and 46 is 23
Answer: 45 players
Step-by-step explanation:
4+3+2=9 9x9=81 9x2=18. 9x3=27 27+18=45
First number is (x).
Second number is (2x+7).
Third number is (x-12).
Their sum is (x+2x+7+x-12).
x+2x+7+x-12 =131
You can simplify this equation
4x-5=131,
4x=136
x=136/4=34 (first number),
(2x+7)=34*2+7=75 (second number)
34-x=34-12 =22(third number)
Check: 34+75+22=131
19/20 in simplest form is 0.95
9/16 in simplest form is 0.5625
