Answer:
Mr. Dairo can produce 20580 pieces of Papaya rosette in two weeks.
Step-by-step explanation:
Papaya rosette pieces on first day = 5890
Papaya rosette Pieces on second day = 7020
Papaya rosette Pieces of third day = 8150
Let
We can check if these numbers are part of a sequence.
In order to check, common difference will be found first.
It can be observed that the common difference is same. When the common difference is same, the sequence is said to be an arithmetic sequence.
The formula for arithmetic sequence is given by:
Putting the values
The formula for nth term can be used to find any term. As we have to find the number of papaya rosette pieces after two weeks which means that we need to find the number of pieces on 14th day.
So,
Hence,
Mr. Dairo can produce 20580 pieces of Papaya rosette in two weeks.
Principal = 1600
annual interest = 6% / year
period = 7 years
Future value after 7 years
= 1600(1+0.06)^7
=1600(1.06^7)
=$2045.81
=$2046 (to the nearest dollar)
Answer:
- 3.75 bags of ChowChow
- 0.75 bags of Kibble
Step-by-step explanation:
The constraints on protein, minerals, and vitamins give rise to the inequalities ...
40c +30k ≥ 150 . . . . . . required protein
20c +20k ≥ 90 . . . . . . required minerals
10c +30k ≥ 60 . . . . . . . required vitamins
And we want to minimize 10c +12k.
The graph shows the vertices of the feasible region in (c, k) coordinates. The one that minimizes cost is (c, k) = (3.75, 0.75).
To minimize cost, the daily feed should be ...
- 3.75 bags of ChowChow
- 0.75 bags of Kibble
Daily cost will be $46.50.
Y - 5 = -2x - 4
Add 5 on both sides.
y = -2x + 1
y = -2x + 1 is in simplest form.
B-8 would be the answer. This is because we are saying you have 8 less than something. So no matter what something is, we are taking away 8 from it. Since our something is unknown, we use our variable 'b' in its place, and subtract 8, giving us b-8.<span />
