4) [tex]$\left(2^{-4}\right)^{-3}$[/tex]

( 10y^3 + 20y^2 +55y ) / 5y = 2y^2 + 4y +11
And then depending on if the equation is set to zero in the problem, just factor or use the quadratic formula on the quadratic equation that is left.

Answered by Kyllie | 2024-06-10

2y^2+4y+11 ;

Answered by zainasayeed1 | 2024-06-12

The simplified form of the expression 5 y 10 y 3 + 20 y 2 + 55 y ​ is 2 y 2 + 4 y + 11 . This is done by factoring out a common term from the numerator and canceling it with the denominator. The final result is a polynomial expression in terms of y .
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Answered by Kyllie | 2024-12-20