Set up a system of equations.
0.10d + 0.25q = 39.25
d + q = 250
Where 'd' represents the number of dimes, and 'q' represents the number of quarters.
d + q = 250
Subtract 'q' to both sides:
d = -q + 250
Plug in '-q + 250' for 'd' in the 1st equation:
0.10(-q + 250) + 0.25q = 39.25
Distribute 0.10:
-0.10q + 25 + 0.25q = 39.25
Combine like terms:
0.15q + 25 = 39.25
Subtract 25 to both sides:
0.15q = 14.25
Divide 0.15 to both sides:
q = 95
Now plug this into any of the two equations to find 'd':
d + q = 250
d + 95 = 250
Subtract 95 to both sides:
d = 155
So there are 95 quarters and 155 dimes.
There are 10 in total but there is 10 and 1/4 total if it has that as a answer
Answer:
c. (x + 3)
Step-by-step explanation:
using factor theorem
if x - 3 is a factor then p(a) = 0
p(a)= x^3 - 3x^2 - 4x + 12
a.(x-3)
p(3) = (3)^3 - 3(3)^2 - 4(3) + 12
= 27 - 27 - 12 + 12
= 0
therefore x-3 is a factor
b.(x + 2)
p(-2) = (-2)^3 - 3(-2)^2 - 4(-2) + 12
= -8 -12 + 8 + 12
,= 0
therefore x + 2 is a factor
c.(x + 3)
p(-3) = (-3)^3 - 3(-3)^2 - 4(-3) + 12
= -27 -27 + 12 + 12
= -30
therefore x + 3 is not a factor
d.(x-2)
p(2) = (2)^3 - 3(2)^2 - 4(2) + 12
= 8 -12 - 8 + 12
= 0
therefore x - 2 is a factor
