He will use 33 containers because 9 goes into 296 32.8 times but you cant have .8 of a container so you round it up to 33
Find the horizontal and vertical distance between the points. First, subtract y2 - y1 to find the vertical distance. Then, subtract x2 - x1 to find the horizontal distance.
An hour and 15 minutes i believe. not 100% though
<span>You can probably just work it out.
You need non-negative integer solutions to p+5n+10d+25q = 82.
If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80.
So this is the same as n + 2d + 5q ≤ 16
So now you simply have to "crank out" the cases.
Case q=0 [ n + 2d ≤ 16 ]
Case (q=0,d=0) → n = 0 through 16 [17 possibilities]
Case (q=0,d=1) → n = 0 through 14 [15 possibilities]
...
Case (q=0,d=7) → n = 0 through 2 [3 possibilities]
Case (q=0,d=8) → n = 0 [1 possibility]
Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81
Case q=1 [ n + 2d ≤ 11 ]
Case (q=1,d=0) → n = 0 through 11 [12]
Case (q=1,d=1) → n = 0 through 9 [10]
...
Case (q=1,d=5) → n = 0 through 1 [2]
Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42
Case q=2 [ n + 2 ≤ 6 ]
Case (q=2,d=0) → n = 0 through 6 [7]
Case (q=2,d=1) → n = 0 through 4 [5]
Case (q=2,d=2) → n = 0 through 2 [3]
Case (q=2,d=3) → n = 0 [1]
Total from case q=2: 1 + 3 + 5 + 7 = 16
Case q=3 [ n + 2d ≤ 1 ]
Here d must be 0, so there is only the case:
Case (q=3,d=0) → n = 0 through 1 [2]
So the case q=3 only has 2.
Grand total: 2 + 16 + 42 + 81 = 141 </span>
Answer:
8 intersections
Step-by-step explanation:
Miss Stevens drove through a total of 36 intersections on her way home from work last week.
At four of every 16 intersections Miss Stevens had to stop for a red light before she could drive through .
This is calculated as:
36 intersections ÷ 16 intersections
= 2 4/16
= 2 1/4
At four of every 16 intersections Miss Stevens had to stop for a red light before she could drive through .
4 of every 16 intersections = stop for a red light
Hence:
16/4 = 4 intersections
Hence: 4 intersections = 1 stop for a read
The number of intersections that Miss Stevens had to stop for a red light is:
2 × 4 intersections = 8 intersections
