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kramer
3 years ago
7

PLEASE HELP!!! SHOW YOUR EXPLANATION!!! Take a rectangular piece of paper. Fold the opposite ends, so that the folding lines are

parallel. How do you know they are parallel?
Mathematics
1 answer:
Alchen [17]3 years ago
5 0
You know they are parallel because the lines will never intersect or meet. (Two straight lines usually mean it is parallel)
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Simplifying expressions help plz
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I got -18-3r 
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3 years ago
Find domain of the rational function<br><br> C(x)=-x-5/x^2-25
iren2701 [21]
\large\begin{array}{l} \textsf{Find the domais of the rational function:}\\\\ \mathsf{C(x)=\dfrac{x-5}{x^2-25}}\\\\\\ \textsf{Denominators can't be zero:}\\\\ \mathsf{x^2-25\ne 0}\\\\ \mathsf{x^2\ne 25}\\\\ \begin{array}{rcl} \boxed{\begin{array}{rcl}\mathsf{x\ne -5}&~\textsf{ and }~&\mathsf{x\ne 5}\end{array}} \end{array} \end{array}


\large\begin{array}{l} \textsf{So the domain of C is}\\\\ \mathsf{D_C=\{x\in\mathbb{R}:~~x\ne -5~~and~~x\ne 5\}}\\\\\\ \textsf{or using a more compact form}\\\\ \mathsf{D_C=\mathbb{R}\setminus\{-5,\,5\}}\\\\\\ \textsf{or in interval notation}\\\\ \mathsf{D_C=\left]-\infty,\,-5\right[\,\cup\,\left]-5,\,5\right[\,\cup\,\left]5,\,+\infty\right[\,.} \end{array}


\large\textsf{So pick up the one you prefer. They're equivalent ways}\\\textsf{to represent the domain.}


If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2184171


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


Tags: domain rational function fraction denominator restriction algebra

4 0
3 years ago
The Lovin lemon company sells a 4 gallon jug of lemonade for $24 dollars. The sweet and Sour company sells an eight pack of 1 qu
aivan3 [116]

Answer:

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sweet & sour co. sells 8 qt = 2 gallons for 16/2 = $8

3 0
3 years ago
20 POINTS HELP Over time, the value of the property at 397 West Lake Street increased by 275%. If the initial value of the prope
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Answer: A is your answer.

6 0
3 years ago
V1 - 2 sin cos 0 = cos 0 - sin​
kramer

\sqrt{1 -  2sin θ \cosθ }  =  \sqrt{ {sin}^{2} θ   + {cos}^{2}θ  -  2 sinθcosθ } =  \sqrt{ {(cosθ   - sinθ) }^{2} }  = cosθ - sinθ

4 0
3 years ago
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