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Andrei [34K]
3 years ago
12

Solve for x b=1/6π n ^2 x

Mathematics
1 answer:
Setler [38]3 years ago
5 0

Answer:

\large\boxed{x=\dfrac{6b}{\pi n^2}}

Step-by-step explanation:

b=\dfrac{1}{6}\pi n^2x\qquad\text{multiply both sides by 6}\\\\6b=\pi n^2x\qquad\text{divide both sides by}\ \pi n^2\neq0\\\\\dfrac{6b}{\pi n^2}=b

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5 Exam-style ABCD is a kite.
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keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of BD

y = \stackrel{\stackrel{m}{\downarrow }}{3}x-1\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}

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(\stackrel{x_1}{-1}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{-\cfrac{1}{3}}[x-\stackrel{x_1}{(-1)}]\implies y-3=-\cfrac{1}{3}(x+1) \\\\\\ y-3=-\cfrac{1}{3}x-\cfrac{1}{3}\implies y=-\cfrac{1}{3}x-\cfrac{1}{3}+3\implies y=-\cfrac{1}{3}x+\cfrac{8}{3}

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2 years ago
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Answer:

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Substitution : m∠1 + m∠7 = 180°

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