$\bf \cfrac{n^2+3n+2}{n^2+6n+8}-\cfrac{2n}{n+4}\implies \cfrac{n^2+3n+21}{(n+4)(n+2)}-\cfrac{2n}{n+4}
\\\\\\
\textit{so our LCD will just be (n+4)(n+2)}
\\\\\\
\cfrac{n^2+3n+2~~~~-~~~~(n+2)(2n)}{(n+4)(n+2)}
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\cfrac{n^2+3n+2~~~~-~~~~(2n^2+4n)}{(n+4)(n+2)}
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\cfrac{n^2+3n+2~~~~-~~~~2n^2-4n}{(n+4)(n+2)}
\implies 
\cfrac{-n^2-n+2}{(n+4)(n+2)}
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\cfrac{-(n^2+n-2)}{(n+4)(n+2)}\implies \cfrac{-\underline{(n+2)}(n-1)}{(n+4)\underline{(n+2)}}\implies \cfrac{-(n-1)}{n+4}\implies \cfrac{1+n}{n+4}$

3 0

2 years ago

Ian made cookies for a bake sale. He made 91 cookies, including 30 cookies in the shape of a dog.

Yuliya22 [10]

Answer:

total sample space is 91

no of cookies in dog shape is 30

therefore

p(d)=30/91

8 0

2 years ago

Denise spent $56.48 at the grocery store while Michael spent $74.06. How much more money did Michael spendthan Denise?A.$17.58B.

ZanzabumX [31]

We will use the equation 74.06-56.48=? To find the change in prices, aka how much more money was spent. That equals 17.58. So A, Michael spent $17.58 more than Denise.

3 0

2 years ago

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