Simplify each expression. Use positive exponents. A^4b^-3 ab^-2

2 answers:

8 0

$\dfrac{a^4b^{-3}}{ab ^{-2}}=\dfrac{a^4}{a^1}\cdot\dfrac{b^{-3}}{b^{-2}}=a^{4-1}\cdot b^{-3-(-2)}=a^3\cdot b^{-3+2}=a^3\cdot b^{-1}=a^3\cdot\dfrac{1}{b}=\\\\\\=\boxed{\dfrac{a^3}{b}}$

choli [55]3 years ago

6 0

$\dfrac{a^4b^{-3}}{ab^{-2}}$

$= {a^{4-1}b^{-3+2}}$

$= {a^{3}b^{-1}}$

$= \dfrac{a^3}{b}$

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Find the equation of a line that is perpendicular to y = 3x – 5 and passes through the point (1, -3).

Svetradugi [14.3K]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5$

well then therefore

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

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

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