Question

Porfavor me ayudan es para hoy

Answer 1

Primero hallamos  "n"

$n\left(n+3\right)=40\\\\n^2+3n=40\\\\\\n^2+3n-40=40-40\\\\\\n^2+3n-40=0\\\\\\n=\frac{-3\ + \sqrt{3^2-4\cdot \:1\cdot \left(-40\right)}}{2\cdot \:1}\\\\\\n=\frac{-3\ + \:13}{2\cdot \:1}\\\\\\n=\frac{-3+13}{2\cdot \:1}\\\\\\n=5$

Luego hallamos el valor "m"

$m\left(2m+1\right)=105\\\\\\2m^2+m=105\\\\\\2m^2+m-105=105-105\\\\\\2m^2+m-105=0\\\\\\m=\frac{-1\ + \sqrt{1^2-4\cdot \:2\left(-105\right)}}{2\cdot \:2}\\\\\\m=\frac{-1\ + \:29}{2\cdot \:2}\\\\\\m=\frac{-1+29}{2\cdot \:2}\\\\\\m=7$

Nos pide hallar  "m.n"

m=7    ,    n=5

7.5=35

Saludos.