Let q be the number of 25 cent coins.
Let d be the number of 10 cent coins.
0.25q+0.10d= 3.95...(1)
q-d=6...(2)
(2)-> q-d= 6
q-d+d= 6+d
q= 6+d...(2a)
(2a)-> (1) 0.25q+0.10d=3.95
0.25(6+d)=0.10d= 3.95
1.95+0.25d+0.10d= 3.95
0.35d= 3.95-1.5
0.35d/0.35= 2.45/0.35
d= 7...(3)
(3)->(2) q-d= 6
q-7= 6
q=6+7
q= 13
There are 13 quarters and 7 dimes.
Answer:
54
Step-by-step explanation:
6x + 5x = 99
11x = 99
x=99/11
x=9
6x = 6x9 = 54
5x = 5x9 = 45
<em>ans: </em><u><em>54</em></u>
Let Brian's steps have a measure of 1, and Richard's steps have a measure of k. Then after each walks 5 steps away from the other, their distance apart is
... 5 + 5k
We are told that distance is equal to 9 of Richard's steps, so is equal to 9k.
... 5 + 5k = 9k
... 5 = 4k . . . . . . . subtract 5k
... 5/4 = k . . . . . . divide by 4
Richard's steps are 5/4 the size of Brian's steps. The appropriate selection is
... b) 5/4
Answer:
The answer is 0.725 Please mark me brainliest xd
You do kcf so 6/2/3 equal to 9
