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KIM [24]
3 years ago
14

Find the missing length indicated. Pls show your work. Thank you

Mathematics
2 answers:
Ksju [112]3 years ago
7 0
<h2>✏️ <u>LENGTH </u> </h2>

\purple{\underline {\bold{\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}

\bold \purple{SOLUTION:3}

» \large \rm \frac{54 - x}{x}  =  \frac{35}{45 - 35}

» \large \rm \frac{54 - x}{x}  =  \frac{35}{10}

» \rm (54 - x )10=  35x

» \rm 540 - 10x=  35x

» \rm 540=  35x + 10x

» \rm 540=  45x

» \large\rm  \frac{540}{45} =   \frac{ \cancel{ \green{45}}x}{ \cancel{ \green{45}}}

» \underline {\rm 12 = x}

\bold \purple{ANSWER:}

» \underline {\boxed {\green{\rm 12 = x}}}

\bold \purple{SOLUTION:4}

» \large\rm  \frac{21}{42}  =  \frac{96 - x}{96 }

» \large\rm  \frac{1}{2}  =  \frac{96 - x}{96 }

» \rm   (96 - x)2 = 96

» \rm   192 - 2x = 96

» \rm    2x= 96   - 192

» \rm    \frac{ \cancel{ \green{ 2}}x}{ \cancel{ \green{ 2}}} = \frac{96}{ -2}

» \underline {\rm x =  48}

\bold \purple{ANSWER:4}

» \underline {\boxed {\green{\rm x = 48}}}

\purple{\underline {\bold{\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}

#LetsStudy

kolbaska11 [484]3 years ago
4 0

#3

\\ \sf\longmapsto \dfrac{54-x}{x}=\dfrac{35}{45-35}

\\ \sf\longmapsto \dfrac{54-x}{x}=\dfrac{35}{10}

\\ \sf\longmapsto 10(54-x)=35x

\\ \sf\longmapsto 540-10x=35x

\\ \sf\longmapsto 540=45x

\\ \sf\longmapsto x=\dfrac{540}{45}

\\ \sf\longmapsto x=12

#4

\\ \sf\longmapsto \dfrac{21}{42}=\dfrac{96-x}{96}

\\ \sf\longmapsto \dfrac{1}{2}=\dfrac{96-x}{96}

\\ \sf\longmapsto 2(96-x)=96

\\ \sf\longmapsto 96-x=48

\\ \sf\longmapsto x=96-48

\\ \sf\longmapsto x=48

Used Law:-

Basic proportionality Theorem.(Thales theorem)

If DE||BC in triangle ABC(D and E are points on AB and AC) then

\\ \sf\longmapsto \dfrac{AD}{DC}=\dfrac{AE}{EC}

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