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mathmath333
 one year ago
Prove
mathmath333
 one year ago
Prove

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} & \normalsize \text{Prove the area of the region enclosed by the graph }\hspace{.33em}\\~\\ & y=x\pm k+5,y=0\hspace{.33em}\\~\\ & \normalsize \text{ is constant for} \ \ k,\ k\in \mathbb{R}\ \hspace{.33em}\\~\\ \end{align}}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I'm finding it hard to visualize... could you show the area in graph ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Ahh that should be easy

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4First of all, we acknowledge that the the absolute value gives the distance between two points on number line: dw:1434119749216:dw

perl
 one year ago
Best ResponseYou've already chosen the best response.0You can prove it using calculus. The integral of that is constant.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4In light that of above thing, \(xk\) represents the distance between \(x\) and \(k\).

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Set the given function equal to \(0\) and solve \(x\) intercepts : \[xk+5=0 \implies xk=5\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yes, subtract them to get the base of triangle

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434120100063:dw
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