Answer:
1. (-5)
Step-by-step explanation:
Here you go, glad to help :)
Answer:
16 minutes
Step-by-step explanation:
<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>
I agree. All real numbers except -1.
An exponential or geometric function can be expressed as a power of t, where t is time.
This means that if you can fit all three values into the formula
S = S0 * (1+r)^t
for a constant r, and t=1, 2, 3 (or 0, 1, 2 for simplicity), then it's exponential.
You can see right away that the first and second sets of numbers are not exponential. These are linear, because each month is a fixed value greater than the previous one.
If you look at the formula above, you can see that each successive time interval's growth can be calculated by multiplying a fixed value to the previous intervals. For example, the second month is given by:
S(1) = S0 * (1+r)
S(2) = S0 * (1+r)^2 = S0 * (1+r) * (1+r) = S(1) * (1+r)
Since each month's sales is 102% the previous month's in the fourth set, this is the one you want.
