Answer:
I think it is D. Because x=x and there's infinite possibilities.
Answer:
How many drinks should be sold to get a maximal profit? 468
Sales of the first one = 345 cups
Sales of the second one = 123 cups
Step-by-step explanation:
maximize 1.2F + 0.7S
where:
F = first type of drink
S = second type of drink
constraints:
sugar ⇒ 3F + 10S ≤ 3000
juice ⇒ 9F + 4S ≤ 3600
coffee ⇒ 4F + 5S ≤ 2000
using solver the maximum profit is $500.10
and the optimal solution is 345F + 123S
Answer:
The value of 1/6 is less than the value of4\5.
So, the answer is < .
Answer:
yes
Step-by-step explanation:
y = 79x - -37
37 = 79(0) - - 37
37 = 0 - - 37
37 = 37
500/20 = 25kcal per piece
8 * 25 = 200kcal.....yes, he is able to have 8 pieces
