Solve the system by substitution. 2.5x−3y=−13, 3.25x−y=−14
We can solve the first equation for x to get x = (3y - 13)/2.5.
Substitute this expression for x into the second equation:
3.25(3y - 13)/2.5 - y = -14
Multiply both sides by 2.5 to eliminate the fractions:
3.25(3y - 13) - 2.5y = -14(2.5)
Distribute and simplify both sides:
9.75y - 42.25 - 2.5y = -35
Combine like terms on the left side:
7.25y - 42.25 = -35
Add 42.25 to both sides:
7.25y = 7.25
Divide both sides by 7.25:
y = 1
Substitute this value of y back into the expression for x:
x = (3(1) - 13)/2.5 = (-10)/2.5 = -4
Therefore, the solution to the system is x = -4 and y = 1.
A factory can produce two products, x and y, with a profit approximated by P=14x 22y−900. The production of y must exceed the production of x by at least 100 units. Moreover, production levels are limited by the formula x 2y≤1400. a.) Identify the vertices of the feasible region. b.) What production levels yield the maximum profit, and what is the maximum profit? Answer a.) and b.)
To identify the vertices of the feasible region, we need to solve the system of inequalities x + 2y ≤ 1400 and y - x ≥ 100.
First, let's rewrite the inequalities as equations by removing the inequality signs:
x + 2y = 1400
y - x = 100
To find the vertices, we solve these equations simultaneously.
We can start by solving the second equation for x:
x = y - 100
Now substitute this expression for x into the first equation:
(y - 100) + 2y = 1400
Combine like terms:
3y - 100 = 1400
Add 100 to both sides:
3y = 1500
Divide both sides by 3:
y = 500
Substitute this value of y back into the expression for x:
x = 500 - 100 = 400
So one vertex is (400, 500).
Now let's solve for the other vertex.
Substitute the expression for x into the equation y - x = 100:
y - (y - 100) = 100
Simplify:
100 = 100
This equation is always true, so there are no restrictions on y. Therefore, y can be any value and x can be determined by the equation x = y - 100.
So a second vertex is (0, 100).
To find the maximum profit, we evaluate the profit function P = 14x + 22y - 900 at each vertex.
At the first vertex (400, 500):
P = 14(400) + 22(500) - 900
P = 5600 + 11000 - 900
P = 16200
At the second vertex (0, 100):
P = 14(0) + 22(100) - 900
P = 0 + 2200 - 900
P = 1300
Therefore, the maximum profit is $16,200 and it occurs at the production levels x = 400 and y = 500.
A factory can produce two products, x and y, with a profit approximated by P=14x+22y−900. The production of y must exceed the production of x by at least 100 units. Moreover, production levels are limited by the formula x+2y≤1400. a.) Identify the vertices of the feasible region. b.) What production levels yield the maximum profit, and what is the maximum profit? Answer a.) and b.)
To identify the vertices of the feasible region, we need to solve the system of inequalities:
x + 2y ≤ 1400
y - x ≥ 100
Let's start by graphing the feasible region.
First, let's rewrite the inequalities as equations by removing the inequality signs:
x + 2y = 1400
y - x = 100
To graph these equations, let's solve for y in terms of x.
From the equation x + 2y = 1400:
2y = 1400 - x
y = (1400 - x)/2
From the equation y - x = 100:
y = x + 100
Now we can graph these two equations:
Plotting the first equation:
x + 2y = 1400
Let's choose some arbitrary x values:
For x = 0, y = 700
For x = 1400, y = 0
Plotting these two points and connecting them with a line, we get a downward sloping line:
Now plotting the second equation:
y - x = 100
Let's choose some arbitrary x values:
For x = 0, y = 100
For x = 1400, y = 1500
Plotting these two points and connecting them with a line, we get an upward sloping line:
The feasible region is the area where both lines intersect, which is the shaded region below:
(1400,0)
(400, 500)
(0,100)
Now let's find the vertices of the feasible region:
- (0, 100)
- (400, 500)
- (1400, 0)
To find the production levels that yield the maximum profit, we substitute the vertices into the profit function P = 14x + 22y - 900, and see which combination gives the highest profit.
For each vertex:
- (0, 100):
P = 14(0) + 22(100) - 900
P = 2200 - 900
P = 1300
- (400, 500):
P = 14(400) + 22(500) - 900
P = 5600 + 11000 - 900
P = 16200
- (1400, 0):
P = 14(1400) + 22(0) - 900
P = 19600 - 900
P = 18700
Therefore, the production levels that yield the maximum profit are x = 400 and y = 500, with a maximum profit of $16,200.
What is the result when 2x^3−9x^2 11x−6 is divided by x−3 ? Show your work.
To divide 2x^3 - 9x^2 + 11x - 6 by x - 3, we can use long division.
2x^2 + -3x + 2
--------------------------------
x - 3 | 2x^3 - 9x^2 + 11x - 6
- (2x^3 - 6x^2)
---------------------------
-3x^2 + 11x
- (-3x^2 + 9x)
-----------------
2x - 6
- (2x - 6)
------------------
0
Therefore, the result of dividing 2x^3 - 9x^2 + 11x - 6 by x - 3 is 2x^2 - 3x + 2.
What is the result when 2x^3−9x^2 +11x−6 is divided by x−3 ? Show your work.
To divide 2x^3 - 9x^2 + 11x - 6 by x - 3, we can use long division.
2x^2 + 3x - 1
---------------------------------
x - 3 | 2x^3 - 9x^2 + 11x - 6
- (2x^3 - 6x^2)
-------------------
-3x^2 + 11x
- (-3x^2 + 9x)
----------------
2x - 6
- (2x - 6)
-----------------
0
Therefore, the result of dividing 2x^3 - 9x^2 + 11x - 6 by x - 3 is 2x^2 + 3x - 1.