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:
307.5
Step-by-step explanation:
He drove 123 miles in 2 hours. If you go 100 miles in 2 hours that means you went 50 miles in one hour right? Using the same logic we can convert 123 miles in 2 hours to 61.5 miles per hour and then multiply it by 5.
Answer:
Step-by-step explanation:
This is already factored for us, which is really nice, so now all we need to do is apply the Zero Product Property to the sets of parenthesis and solve for x, which will give us the 2 times that the object is on the ground.
x + 1 = 0 so
x = -1
x - 9 = 0 so
x = 9
We all know that time cannot ever be negative, so the time that the object is on the ground is 9 seconds after it's launched (which was from an initial height of 45 meters).
3y = 5x + 30
y + 5x = 50 .......(1)
3y - 5x = 30 .....(2) - rearranging the first equation.
Add (1) and (2):-
4y = 80
y = 20
Now plug y = 20 into equation (1):-
20 + 5x = 50
5x = 30
x = 6
The 2 numbers are 6 and 20.