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asambeis [7]
4 years ago
10

How to simplify this expression

Mathematics
1 answer:
WITCHER [35]4 years ago
4 0
\frac{2x+4}{3x-3}( \frac{12x-12}{x+5})
=\frac{2(x+2)}{3(x-1)}( \frac{12(x-1)}{x+5})
=\frac{2(x+2)4}{x+5}
=\frac{8(x+2)}{x+5}

--------------------------------------
\frac{x^2-16}{2x-8}÷ \frac{x-4)^2}{8x-32}
=\frac{x^2-16}{2x-8}( \frac{8x-32}{x-4)^2})
=\frac{(x-4)^2}{2x-8}( \frac{8x-32}{x-4)^2})
=\frac{1}{2x-8}( \frac{8x-32}{1})
=\frac{8x-32}{2x-8}
=\frac{8(x-4)}{2(x-4)}
=\frac{4(x-4)}{1(x-4)}
=4
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5/6 (x - 1) = 4 what Is x
vfiekz [6]
5/6(x-1)=4
(x-1)=(4*6)/5
x-1=24/5
x=24/5  +  1
least common multiple=5
x=(24+1*5)/5
x=29/5

Answer: x=29/5

To check:

5/6(x-1)=5/6(29/5  -1)=5/6[(29-5)/5]=5/6(24/5)=(5*24)/(6*5)=120/30=4

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3 years ago
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4.
Svetlanka [38]

Step-by-step explanation:

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3 years ago
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:

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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 ...

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subject to the constraints on drape production:

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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

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

_____

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.

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