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Svetlanka [38]
3 years ago
12

The local jewelry store marks up the price of a necklace by 65% before selling them. If the store paid the distributor $85 per n

ecklace how much will a necklace cost in the store
Mathematics
1 answer:
Zolol [24]3 years ago
6 0
In the store, the necklace will cost $140.25. This is because we are taking the price of 85$ and adding on an extra 65% of it to the number.

85(1.65) = 140.25
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andrey2020 [161]
To solve both of these expressions, we want to add or subtract from left to right.

For the first one, we can rewrite it as:

8-6-9
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The second one we can rewrite as:

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5 0
3 years ago
What is true about the solution of x^2/2x-6=9/6x-18?
pogonyaev

Answer:

  • x = ±√3, and they are actual solutions
  • x = 3, but it is an extraneous solution

Step-by-step explanation:

The method often recommended for solving an equation of this sort is to multiply by the product of the denominators, then solve the resulting polynomial equation. When you do that, you get ...

... x^2(6x -18) = (2x -6)(9)

... 6x^2(x -3) -18(x -3) = 0

...6(x -3)(x^2 -3) = 0

... x = 3, x = ±√3

_____

Alternatively, you can subtract the right side of the equation and collect terms to get ...

... x^2/(2(x -3)) - 9/(6(x -3)) = 0

... (1/2)(x^2 -3)/(x -3) = 0

Here, the solution will be values of x that make the numerator zero:

... x = ±√3

_____

So, the actual solutions are x = ±3, and x = 3 is an extraneous solution. The value x=3 is actually excluded from the domain of the original equation, because the equation is undefined at that point.

_____

<em>Comment on the graph</em>

For the graph, we have rewritten the equation so it is of the form f(x)=0. The graphing program is able to highlight zero crossings, so this is a convenient form. When the equation is multiplied as described above, the resulting cubic has an extra zero-crossing at x=3 (blue curve). This is the extraneous solution.

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netineya [11]

Answer:

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Step-by-step explanation:

7 0
3 years ago
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#6, #7, and #10 please:)
scoundrel [369]
All given xd ur welcome so ez
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