Let us say that:
a = ones
b = fives
c = twenties
So that the total money is:
1 * a + 5 * b + 20 * c = 229
=> a + 5b + 20c = 229 -->
eqtn 1
We are also given that:
c = a – 5 -->
eqtn 2
a + b + c = 30 -->
eqtn 3
Rewriting eqtn 3 in terms of b:
b = 30 – a – c
Plugging in eqtn 2 into this:
b = 30 – a – (a – 5)
b = 35 – 2a -->
eqtn 4
Plugging in eqtn 2 and 4 into eqtn 1:
a + 5(35 – 2a) + 20(a – 5) = 229
a + 175 – 10a + 20a – 100 = 229
11a = 154
a = 14
So,
b = 35 – 2a = 7
c = a – 5 = 9
Therefore there are 14 ones, 7 fives, and 9 twenties.
2/3 would be the answer hope this helps.
Answer:
B
Step-by-step explanation:
Expressing the ratios as fractions, that is
=
( cross- multiply )
a = 3b
=
( cross- multiply )
c = 5b
------------------------------------------
Given
, substitute values from above for a and c
= 
= 
=
← cancel b on numerator/ denominator
=
→ B
Answer:
A
Step-by-step explanation:
The line cuts the y at positive two. The rise over run of the slope is -3/1, or 3.
Just had a test on these kinds of problems and i got 100.