Simplify the complex fraction. y-1/y^2+y-6/y-6/y+3

2 answers:

8 0

For this case we have the following complex fraction:
$\frac{\frac{y-1}{y^2+y-6} }{\frac{y-6}{y+3}}$
Using the double C method, we can rewrite the given fraction.
We have then:
$\frac{(y-1)(y+3)}{(y^2+y-6)(y-6)}$
Then, we must factor the quadratic expression into the denominator.
We have then:
$\frac{(y-1)(y+3)}{(y-2)(y+3)(y-6)}$
Finally, we cancel similar terms.
We have then:
$\frac{(y-1)}{(y-2)(y-6)}$
Answer:
the complex fraction simplified is:
B) y-1/(y-6)(y-2)

otez555 [7]4 years ago

3 0

Answer:

Step-by-step explanation:

try to do a or b

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The Janie Gioffre Drapery Company makes three types of draperies at two different locations. At location I, it can make 10 pairs

Tcecarenko [31]

Answer:

30 days at location I
35 days at location II

Step-by-step explanation:

Let x and y represent days of operation of Location I and Location II, respectively. Then we want to minimize the objective function ...

650x +750y

subject to the constraints on drape production:

10x +20y ≥ 1000 . . . . . order for deluxe drapes

20x +50y ≥ 2100 . . . . . order for better drapes

13x +6y ≥ 600 . . . . . . . order for standard drapes

I find a graphical solution works well for this. The vertices of the feasible solution space are (x, y) = (0, 100), (30, 35), (80, 10), (105, 0). The vertex at which the cost is minimized is

(x, y) = (30, 35)

This schedule will produce exactly the required numbers of deluxe and standard drapes, and 2350 pairs of better drapes, 250 more than required.

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In the attached graph, we have reversed the inequalities so that the solution space (feasible region) is white, not triple-shaded. Minimizing the objective function means choosing the vertex of the feasible region so that the line representing the objective function is as close to the origin as possible.