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krek1111 [17]
3 years ago
9

The length of a rectangle is two thirds its width. the area is 12 square meters. find its width. round your answer to two decima

l places.
Mathematics
1 answer:
ziro4ka [17]3 years ago
5 0
L = ( \frac{2}{3} )w
a = 12 
Length times width = area 
L · w = a
( \frac{2}{3} )w · w = 12
( \frac{2}{3} ) w^{2} = 12
Divide both sides by \frac{2}{3} or mulitply both sides by \frac{3}{2} (either one is fine)
( \frac{2}{3} ) ÷ ( \frac{2}{3} ) w^{2} = 12 ÷ ( \frac{2}{3} )
or
( \frac{3}{2} ) ( \frac{2}{3} ) w^{2} = 12 ( \frac{3}{2} )
w^{2} = 18
Square root both sides to get rid of the squared on the w
\sqrt{w^{2} } = \sqrt{18}
w = 4.24
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<h3>Given Equation:-</h3>

\boxed{ \rm  \frac{4x^{2}y^{3}z}{9} \times  \frac{45y}{8 {x}^{5} {z}^{5} }}

<h3>Step by step expansion:</h3>

\dashrightarrow \sf\dfrac{4x^{2}y^{3}z}{9} \times  \dfrac{45y}{8 {x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{ \cancel4x^{2}y^{3}z}{9} \times  \dfrac{45y}{ \cancel8 {x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{9} \times  \dfrac{45y}{2{x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{ \cancel9} \times  \dfrac{ \cancel{45}y}{2{x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{1} \times  \dfrac{5y}{2{x}^{5} {z}^{3} }

\\  \\

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\\  \\

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\\  \\

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\\  \\

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\\  \\

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\\  \\

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\therefore \underline{ \textbf{ \textsf{option \red c \: is \: correct}}}

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