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solmaris [256]
3 years ago
10

2/4 - 1/4 please hep i know im not that good for asking this

Mathematics
1 answer:
Nady [450]3 years ago
3 0
Since they have equal bases you can subtract it like a regular number. 2-1 equals 1 so the answer would be 1/4
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(x+7)/(x^2-49) find the domain. show work
harina [27]
\large\begin{array}{l} \textsf{Find the domain of}\\\\ \mathsf{f(x)=\dfrac{x+7}{x^2-49}}\\\\ \mathsf{f(x)=\dfrac{x+7}{x^2-7^2}}\\\\\\ \textsf{Factor out the denominator using special products:}\\\\ \textsf{(a difference of squares)}\\\\ \mathsf{f(x)=\dfrac{x+7}{(x+7)(x-7)}} \end{array}


\large\begin{array}{l} \textsf{Restrictions for the domain:}\\\\ \bullet~~\textsf{Denominators must not be zero:}\\\\ \mathsf{(x+7)(x-7)\ne 0}\\\\ \begin{array}{rcl} \mathsf{x+7\ne 0}&~\textsf{ and }~&\mathsf{x-7\ne 0}\\\\ \mathsf{x\ne -7}&~\textsf{ and }~&\mathsf{x\ne 7} \end{array} \end{array}


\large\begin{array}{l} \textsf{Therefore, the domain of f is}\\\\ \mathsf{D_f=\{x\in\mathbb{R}:~~x\ne -7~~and~~x\ne 7\}}\\\\\\ \textsf{or using a more compact form}\\\\ \mathsf{D_f=\mathbb{R}\setminus\{-7,\,7\}}\\\\\\ \textsf{or using the interval notation}\\\\ \mathsf{D_f=\left]-\infty,\,-7\right[\,\cup\,\left]7,\,+\infty\right[.} \end{array}


<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2155752


\large\textsf{I hope it helps. :-)}



Tags: <em>function domain real rational factorizing special product interval</em>

</span>
7 0
3 years ago
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