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laila [671]
3 years ago
14

Susan enlarged a rectangle with a height of 6 cm and length of 13 cm on her computer. The length of the new rectangle is 19.5 cm

. Find the height of the new rectangle.
Mathematics
1 answer:
Karolina [17]3 years ago
4 0
We have that enlarging the rectangle in a computer setting keeps the ratio of the lines constant. Hence, height over length is the same. For the first rectangle this is 6/13. For the second one, if x is the new height, this is x/19.5.
Hence x/19.5=6/13. Multiplying both parts by 19.5 we have that x=19.5*6/13=9. Thus, the new height is 6 cm.
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The line y =3x-5 meet x-axis at the point M.
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3y + 2x = 2, meets the y-axis when x = 0, and that'd be the y-intercept


\bf 3y+2(\stackrel{x}{0})=2\implies 3y=2\implies y=\cfrac{2}{3}~\hspace{5em}\stackrel{N}{\left( 0,\frac{2}{3} \right)}


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\bf M(\stackrel{x_1}{\frac{5}{3}}~,~\stackrel{y_1}{0})\qquad N(\stackrel{x_2}{0}~,~\stackrel{y_2}{\frac{2}{3}}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{~~\frac{2}{3}-0~~}{0-\frac{5}{3}}\implies \cfrac{2}{3}\cdot -\cfrac{3}{5}\implies -\cfrac{2}{5}


\bf \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-0=-\cfrac{2}{5}\left( x-\cfrac{5}{3} \right)\implies y=-\cfrac{2}{5}x+\cfrac{2}{3} \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{15}}{15y=-6x+10}\implies \stackrel{\textit{standard form}}{6x+15y=10}\implies 6x+15y-10=0

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3 years ago
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