Respuesta :
h(x)=[tex]-10x^{2}+40x[/tex]
Step-by-step explanation:
given ,[tex]f(x)=2x-8[/tex]
given, [tex]g(x)=-5x[/tex]
Given that [tex]h(x)=f(x)g(x)[/tex]
Let [tex]t_{1}[/tex] be [tex]k_{1}x^{n_{1}}[/tex] where [tex]k_{1} \text{ and } n_{1} \text{ are constants}[/tex]
Let [tex]t_{2}[/tex] be [tex]k_{2}x^{n_{2}}[/tex] where [tex]k_{2} \text{ and } n_{2} \text{ are constants}[/tex]
We know that [tex]t_{1}\times t_{2}=k1\times x^{n_{1}}\times k2\times x^{n_{2}}[/tex]=[tex]k_1k_2x^{n_{1}+n_{2}}[/tex]
So,[tex]h(x)=(2x-8)(-5x)[/tex]=[tex]-10x^{2}+40x[/tex]
Answer:
B : h(x) = –10x2 + 40x
Step-by-step explanation:
this is a late answer but I hope it helps anyone who needs it for the future !!