Question

pls solve this for me!!!
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Answer 1

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$iv ) \: \: \frac{5xy + 3z}{2a {}^{3} - c {}^{2} }$

When x =2 , y = -1, z = 3 , a = 4 and c = 1/3

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$= > \frac{5 \times 2 \times ( - 1) + 3 \times 3}{2(4) {}^{3} - ( \frac{1}{3} ){}^{2} }$

$= > \frac{ - 10 + 9}{2 \times 64 - \frac{1}{9} }$

$= > \frac{ - 1}{128 - \frac{1}{9} }$

$= > \frac{ - 1}{ \frac{1152 - 1}{9} }$

$= > \frac{ - 1}{ \frac{1151}{9} }$

$= > \frac{ - 1 \times 9}{1151}$

$= > \frac{ - 9}{1151}$

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Answer 2

Answer:

➡ $\sf \frac{-9}{1151}$

Step-by-step explanation:

Given fraction :

$\sf\longrightarrow \frac{5xy + 3z}{{2a}^{3} - {c}^{2}}$

Where the given values are :

➡ x = 2 ; a = 4

➡ y = ( - 1) ; c = $\sf \frac{1}{3}$

➡ z = 3

Substituting the given values :

$\sf \longrightarrow \: \frac{5xy + 3z}{ {2a}^{3} - {c}^{3} }$

$\: \sf \longrightarrow \: \frac{5(2)( - 1) + 3(3)}{ {2(4)}^{3} - ( { \frac{1}{3}) }^{2} }$

$\: \sf \longrightarrow \: \frac{ - 10 + 9}{128 - \frac{1}{9} }$

$\sf \longrightarrow\: \frac{ - 1}{ \frac{1152 - 1}{9} }$

$\sf \longrightarrow \: \frac{ - 1}{ \frac{1151}{9} }$

$\sf \longrightarrow \frac{ - 1}{1151} \times 9$

$\sf \longrightarrow \: \frac{ - 9}{1151} \: ans.$