Subtract 7000 and 1330 which gives you 5670 then you want to simplify and get 19.0
Let "c" and "q" represent the numbers of bottles of Classic and Quantum that should be produced each day to maximize profit. The problem conditions give rise to 3 inequalities:
.. 0.500c +0.550q ≤ 100 . . . . . . . liters of water
.. 0.600c +0.200q ≤ 100 . . . . . . . kg of sugar
.. 0.1c +0.2q ≤ 32 . . . . . . . . . . . . . grams of caramel
These can be plotted on a graph to find the feasible region where c and q satisfy all constraints. You find that the caramel constraint does not come into play. The graph below has c plotted on the horizontal axis and q plotted on the vertical axis.
Optimum production occurs near c = 152.17 and q = 43.48. Examination of profit figures for solutions near those values reveals the best result for (c, q) = (153, 41). Those levels of production give a profit of 6899p per day.
To maximize profit, Cartesian Cola should produce each day
.. 153 bottles of Classic
.. 41 bottles of Quantum per day.
Profit will be 6899p per day.
_____
The problem statement gives no clue as to the currency equivalent of 100p.
Answer:
18%
Step-by-step explanation:
Multiply 22.50 by all the answer choices and use PROCESS OF ELIMINATION.
By default 18 percent is what gives you the answer 4.05
Answer:
![y = 5(6) {}^{x}](https://tex.z-dn.net/?f=y%20%3D%205%286%29%20%7B%7D%5E%7Bx%7D%20)
Step-by-step explanation:
A exponential function is represented by
![y = ab {}^{x}](https://tex.z-dn.net/?f=y%20%3D%20ab%20%7B%7D%5E%7Bx%7D%20)
where a is the vertical stretch and b is the base and x is the nth
power of x
Since the first number corresponds with zero, that means our y intercept is the first number.
This means when x=0 , y=5 so let find the value of 5.
![5 = ab {}^{0}](https://tex.z-dn.net/?f=5%20%3D%20ab%20%7B%7D%5E%7B0%7D%20)
b to the 0th power equal 1 so
![5 = a \times 1](https://tex.z-dn.net/?f=5%20%3D%20a%20%5Ctimes%201)
![a = 5](https://tex.z-dn.net/?f=a%20%3D%205)
Our equation is for now
![y = 5b {}^{x}](https://tex.z-dn.net/?f=y%20%3D%205b%20%7B%7D%5E%7Bx%7D%20)
Now let plug in 1,30
![30 = 5b {}^{1}](https://tex.z-dn.net/?f=30%20%3D%205b%20%7B%7D%5E%7B1%7D%20)
Divide 5 by both sides
![6 = b {}^{1}](https://tex.z-dn.net/?f=6%20%3D%20b%20%7B%7D%5E%7B1%7D%20)
Anything to the 1st power is itself so b equal 6.
So our equation is
![y = 5(6) {}^{x}](https://tex.z-dn.net/?f=y%20%3D%205%286%29%20%7B%7D%5E%7Bx%7D%20)