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dlinn [17]
3 years ago
7

What is the difference in simplest form?

Cfrac%7B2n%7D%7Bn%2B4%7D%20%20%20" id="TexFormula1" title=" \frac{n^2+3n+2}{n^2+6n+8}-\frac{2n}{n+4} " alt=" \frac{n^2+3n+2}{n^2+6n+8}-\frac{2n}{n+4} " align="absmiddle" class="latex-formula">
First picture example problem
Second picture problem with the answers.

Mathematics
1 answer:
Paha777 [63]3 years ago
3 0
\bf \cfrac{n^2+3n+2}{n^2+6n+8}-\cfrac{2n}{n+4}\implies \cfrac{n^2+3n+21}{(n+4)(n+2)}-\cfrac{2n}{n+4}
\\\\\\
\textit{so our LCD will just be (n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(n+2)(2n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(2n^2+4n)}{(n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~2n^2-4n}{(n+4)(n+2)}
\implies 
\cfrac{-n^2-n+2}{(n+4)(n+2)}
\\\\\\
\cfrac{-(n^2+n-2)}{(n+4)(n+2)}\implies \cfrac{-\underline{(n+2)}(n-1)}{(n+4)\underline{(n+2)}}\implies \cfrac{-(n-1)}{n+4}\implies \cfrac{1+n}{n+4}
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Step-by-step explanation:

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The slope of the line below is -2. Use the coordinates of the labeled point to
Vilka [71]

Answer:

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Step-by-step explanation:

Incomplete question:

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We can express the equation for any linear equation with slope -2 and point (x0,y0) as:

y=-2x+(y_0+2x_0)\\\\y-y_0=-2(x-x_0)

6 0
3 years ago
In the diagram below, P is circumscribed about quadrilateral ABCD. What is the value of x?
alexira [117]
We know that
In any quadrilateral where the vertices are on the circumference of a circle, the opposite angles add up to 180 degrees. 

So,
 in this problem
(x+20) + 110 = 180 \\ x+130=180 \\ x=180-130 \\ x=50degrees

therefore

the answer is
x=50degrees
7 0
3 years ago
Read 2 more answers
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