Answer:
P(t) = 100t +2600
4100 in 2005
Step-by-step explanation:
You are given two points for (year, population) = (t, p):
(4, 3000), (8, 3400)
It is useful to use the two-point form of the equation for a line.
p = (p2 -p1)/(t2 -t1)(t -t1) +p1
p = (3400 -3000)/(8 -4)(t -4) +3000
p = 400/4(t -4) +3000
p = 100t +2600
P(t) = 100t +2600 . . . . written in functional form
In 2005, the population is predicted to be ...
P(15) = 100×15 +2600 = 4100
Answer:
Ask teacher
Step-by-step explanation:
Answer:
coordinates will be (1/2 , -5)
Step-by-step explanation:
when x=4
y= -2(4)+3
y = -8 +3
y = -5
when y=2
2 = - 2x + 3
2-3 = -2x
-1 = -2x
1=2x
x = 1/2
Answer:
a) Minimize 
subject to




b) Attached
c) The optimum value that minimizes cost is x1=28 and x2=8.
Step-by-step explanation:
The objective function is the cost of extraction and needs to be minimized.
The cost of extraction is the sum of the cost of extraction of ore type 1 and the cost of extraction of ore type 2:

Being x1 the tons of ore type 1 extracted and x2 the tons of ore type 2.
The constraints are the amount of minerals that need to be in the final mix
Copper:

Zinc

Magnesium

Of course, x1 and x2 has to be positive numbers.

The feasible region can be seen in the attached graph.
The orange line is the magnesium constraint. The red line is the copper constraint. The green line is the zinc constraint.
The optimal solution is found in one of the intersection points between two constraints that belong to the limits of the feasible region.
In this case, the cost can be calculated for the 3 points that satisfies the conditions.
The optimum value that minimizes cost is x1=28 and x2=8.