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dimaraw [331]
3 years ago
15

How do I solve this

Mathematics
1 answer:
anastassius [24]3 years ago
7 0
\bf \cfrac{x+2}{x^2+6x-7}    so, that function is "defined", ok, what values of "x" are not in the domain, namely, what values can "x" take on and not make the function "undefined", well,  you know, if we end up with a 0 at the denominator, like   \bf \cfrac{x+2}{0}    then, we'd have an "undefined" expression...so... any values of "x" that make the denominator 0, are not really the ones we want, and thus they'd be excluded from the domain.


so, hmm which are those? let's check, let's set the denominator to 0, and solve for "x".

\bf x^2+6x-7=0\implies (x+7)(x-1)=0\implies x=
\begin{cases}
-7\\
1
\end{cases}
\\\\\\
\textit{let's check, } x=-7\quad \cfrac{(-7)+2}{(-7)^2+6(-7)-7}\implies \cfrac{-5}{49-42-7}\implies \cfrac{-5}{0}
\\\\\\
x=1\quad \cfrac{(1)+2}{(1)^2+6(1)-7}\implies \cfrac{3}{1+6-7}\implies \cfrac{-3}{0}
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This is kind of late, but I hope I could help maybe for a later assignment or assesment.

The estimates of y have become increasingly more precise, with some of the earlier ones having a margin of error of more than 0.8

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ElenaW [278]

Hello from MrBillDoesMath!

Answer:

n = 5


Discussion:

Let the original number be "n"  Translating the problem  statement to mathematical symbolism:

n + 4 = 19 - 2n                   =>  add 2n to both sides

n + 2n + 4 = 19 - 2n + 2n  =>  combine like terms

3n + 4 = 19                        => subtract 4 from both sides

3n = 19 -4  = 15                 => divide both sides by 3

n = 15/3 =5


Check: Does 5 +4 = 19 - 2(5)? Does 9 = 9? Yes. so the value for n is confirmed correct.


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