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
1.Not Even
2.Even
3.Not Even
4.Not Even
X^2-2x -3 =0
a =1 b =-2 and c = -3
x = - (-2) +/- sqrt (-2)^2 - 4(1)(-3)
--------------------------------------
2(1)
x = + 2 +/- sqrt [(4) - 4(-3)]
-------------------------------
2
x = +2 +/- sqrt [4 + 12]
--------------------------
2
x = +2 +/- sqrt[16]
-------------------
2
x = +2 +/- (4)
-------------
2
x = 2 + 4 or 2 -4
------- ------
2 2
x = 6/2 or -2/2
x = 3 or x = -1
Answer:
d
Step-by-step explanation:
just makes sense bc they're all in 2nd grade
