Question 407367

1.   {{{ b > 11.2 }}}...{{{b}}} cold be any number greater than {{{11.2}}}, like

{{{b= 11.21}}}, {{{b= 11.22}}},...{{{b= 11.3}}}...and so on 


2.   

{{{x(7-x)>8}}}

{{{7x-x^2>8}}}

{{{7x-x^2-8>0}}}...solve quadratic as it is equal to {{{0}}}


{{{7x-x^2-8=0}}}

{{{-x^2+7x-8=0}}}


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-7 +- sqrt( 7^2-4*(-1)*(-8) ))/(2*(-1)) }}}

{{{x = (-7 +- sqrt( 49-32 ))/(-2) }}}

{{{x = (-7 +- sqrt( 17 ))/(-2) }}}

{{{x = (-7 +- ( 4.12))/(-2) }}}


{{{x1 = (-7 + 4.12)/(-2) }}}

{{{x1 = (-2.88)/(-2) }}}

{{{x1 = 1.44 }}}

{{{x2 = (-7 - 4.12)/(-2) }}}

{{{x2 = (-11.12)/(-2) }}}

{{{x2 = 5.56 }}}


{{{x(7-x)>8}}}.........plug in {{{x1}}} and {{{x2}}} to see what will be a solution for this inequality


{{{x(7-x)>8}}}...{{{1.44(7-1.44)>8}}}....{{{1.44(5.56)>8}}}...{{{8.0064
>8}}}

for {{{x2 = 5.56 }}}
{{{5.56(7-5.56)>8}}}...{{{5.56(7-5.56)>8}}}....{{{1.44(1.44)>8}}}...{{{2.0736
>8}}}...this is not a solution

so, your inequality will be true only if 

{{{x = 1.44 }}}



3.    {{{6 <= 6k}}}

{{{6/6 <= k}}}

{{{1 <= k}}}...{{{k}}} could be {{{1}}} and any number greater than {{{1}}}