If $p=3, q=-5$ and $r=-1$ , what is $(5p+q)/(q-8r) + (4q^2)/(a+5p)$ ?

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Answer 1 · 2017-05-03T08:07:39

$(10/3)+(100/(a+15))$

If you assign these values into the equation:

$(5*3-5)/(-5-(8*-1)) + (4*(-5^2))/(a+5*3)$

It will be:

$((15-5)/(-5+8)) + (100/(a+15))$

$10/3 + 100/(a+15)$

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$10/3 + 100/(a+15)$

Answer 2 · 2017-05-03T08:11:29

$(5p+q)/(q-8r)+(4q^2)/(a+5p)=(10a+450)/(3a+15)$

As $p=3$ , $q=-5$ and $r=-1$

$(5p+q)/(q-8r)+(4q^2)/(a+5p)$

= $(5xx3+(-5))/((-5)-8(-1))+(4(-5)^2)/(a+5xx3)$

= $(15-5)/((-5)+8)+(4xx25)/(a+15)$

= $10/3+100/(a+15)$

= $(10(a+15)+100xx3)/(3(a+5))$

= $(10a+150+300)/(3(a+5))$

= $(10a+450)/(3a+15)$

If p=3, q=-5p=3, q=-5 and r=-1r=-1, what is (5p+q)/(q-8r) + (4q^2)/(a+5p)(5p+q)/(q-8r) + (4q^2)/(a+5p)?