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yarga [219]
3 years ago
6

!!!!!!!!!!!!!!!!!!!!!!!!!! Plz

Mathematics
1 answer:
gayaneshka [121]3 years ago
4 0

The answer is c option which is 6.7 units

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Please Help! I need an explanation for how to do this... Step-by-step directions would be great!
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\dfrac{8x^2y^2-4xy^2}{4xy}=\\
\dfrac{8x^2y^2}{4xy}-\dfrac{4xy^2}{4xy}=\\
2xy-y

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Describe how to find the selling price of an item that has been marked up 110%
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Step-by-step explanation:

The original price on the item is the price before a markup of 110% of this original price:

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3 years ago
Que pasa si un objeto imparable choca con uno inabobible​
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English Answer: No object can resist an irresistible force. No force can move an immovable object. So if an immovable object meets an irresistible force it will move and not move.English:

Spanish Answer: Ningún objeto puede resistir una fuerza irresistible. Ninguna fuerza puede mover un objeto inamovible. Entonces, si un objeto inamovible se encuentra con una fuerza irresistible, se moverá y no se moverá.

7 0
2 years ago
Read 2 more answers
PLEASE HELP ITS URGENT
Sveta_85 [38]
X^2-22x-48=0
x^2-24x+2x-48=0
x(x-24)+2(x-24)
(x+2)(x-24)

Solve by grouping if you are able to find distinct factors that multiply to the last term and add to the middle term...this method is rather easy with easy to manage numbers. Complete the square if you cannot find distinct factors that multiply to the last term and add to the middle term. Completing the square helps when the equation is in the form of a parabola.
6 0
3 years ago
I have the first part but I don't know how to do the second part<br><br> h e l p
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\bf \begin{array}{cll} \stackrel{ordinal}{position}&\stackrel{term}{value}\\ \cline{1-2} 1&12\\ 2&12+d\\ 3&12+d+d\\ 4&12+d+d+d\\ &12+3d \end{array}\qquad \implies ~\hfill \begin{array}{llll} \stackrel{\textit{4th term}}{12+3d}~~=~~\stackrel{\textit{4th value}}{3}\\\\ 3d=-9\implies d = \cfrac{-9}{3}\\\\ \stackrel{\textit{common difference}}{d = -3} \end{array}

well, we know the common difference is -3, to go from the 4th term to the 8th term, we need to add "d" 4 times or namely 3+4(-3), likewise to go from the 13th term to the 19th term we have to add "d" 6 times, or namely -24 + 6(-3).

\bf \stackrel{\textit{8th term}}{3+4(-3)}\implies 3-12\implies -9 \\\\\\ \stackrel{\textit{19th term}}{-24 + 6(-3)}\implies -24-18\implies -42

3 0
2 years ago
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