3 years ago

I reallyyy need helppp

Mathematics

1 answer:

Julli [10]3 years ago

6 0

Cool ok so chat me and I'll help

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Solve the LP problem. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is

Ksivusya [100]

Answer:

Step-by-step explanation:

Minimize c = -x + 5y

The constraints say

2x >= 3y, x<=3y, y>=4 and x>=6, x+y<=12

Since we need to minimize y and maximize x in order to minimize c

y_(min) = 4

x_(max) <= 3y_(min) <= 12

which is also a constraint from x + y <= 16

Hence the closest feasible solution will be (12,4)

Therefore, minimum value of c will be -12 + 5(4) = 8

Hence the final answer is equal to 8

8 0

4 years ago

Emily is buying new cold-weather gear for her family. She pays $100 for 5 pairs

user100 [1]

Answer:

500, or 400 if its 4

Step-by-step explanation:

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3 years ago

What is the area, in square units, of the parallelogram shown below? A parallelogram ABCD is shown with height 8 units and base

GenaCL600 [577]

Answer:

Step-by-step explanation:

Area of parallelogram = base * height

= 6 * 8

= 48 square units

5 0

3 years ago

Read 2 more answers

Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

$\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)$

$f'(y)=0\implies f(y)=C$

$\implies F(x,y)=x^2y^3+\ln(x+y^2)=C$

With $y(1)=2$ , we have

$8+\ln9=C$

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

8 0

3 years ago

the pressure, p. of a gas varies inversely with its volume, v. Pressure is measured in units of Pa. Suppose that a particular am

LekaFEV [45]

I think I know how to do it

Your equation is:

P₁V₁ = P₂V₂

You Know:

Initial:

P₁ = 84 Pa

V₁ = 336 L

New:

P₂ = ?

V₂ = 216L

Plug this into the equation:

P₁V₁ = P₂V₂

(84 Pa)(336 L) = P₂(216 L) Divide 216 on both sides

$\frac{(84Pa)(336L)}{216L}=\frac{(P_{2})(216L)}{216L}$

130.6666666 = P₂

(if your doing significant figures, you have 2 sig figs)

130 Pa = P₂

6 0

3 years ago