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
I didn’t see an image or anything but I looked it up and it said regular polygons, since they have the sides all the same length they must always be in the same proportions, and their interior angles are always the same.
F(g(x)) = [(-7x-8)/(x-1) - 8} / [(-7x - 8)/(x-1) + 7] =
[(-7x - 8 - 8(x-1)) / (x-1)] / [(-7x - 8 + 7(x-1)) / (x-1)] = (-15x) / (-15) = x.
g(f(x)) = [-7*(x-8)/(x+7) - 8] / [(x-8)/(x+7) - 1] =
[(-7x + 56 -8*(x+7)) / (x+7)] / [(x - 8 - (x + 7)) / (x+7)] = (-15x) / (-15) = x.
So since f(g(x)) = g(f(x)) = x we can conclude that f and g are inverses.
.1 Simplify: 5n(2n3+n2+8)+n(4-n).
Solution:
5n(2n3+n2+8)+n(4-n).
= 5n × 2n3 + 5n × n2 + 5n × 8 + n × 4 - n × n.
= 10n4 + 5n3 + 40n + 4n – n2.
= 10n4 + 5n3 + 44n – n2.
= 10n4 + 5n3 – n2 + 44n.
Answer: 10n4 + 5n3 – n2 + 44n
The seller makes a over all loss of $2000
