A .[tex]Length = x+5\\breadth = x-2[/tex]
B. domain of [tex]f(x)= x^2+3x-10[/tex] here is all values of x > 2 .
Step-by-step explanation:
Correct question is : the function [tex]f(x)= x^2+3x-10[/tex]models the area of a rectangle. a: describe the length and width of the rectangle in terms of x b: what is a domain for above function in relation to rectangle . Let's solve:
a: describe the length and width of the rectangle in terms of x
Let's factorise [tex]f(x)= x^2+3x-10[/tex] :
⇒ [tex]f(x)= x^2+5x-2x-10[/tex]
⇒ [tex]f(x)= x(x+5)-2(x+5)[/tex]
⇒ [tex]f(x)= (x+5)(x-2)[/tex]
Since , ⇒ [tex]f(x)=length(breadth)[/tex] So
[tex]Length = x+5\\breadth = x-2[/tex] .
b: what is a domain for above function in relation to rectangle
Since area is always > 0 , and (length , breadth)>0 So,
⇒ [tex]f(x)= (x+5)(x-2)>0\\x>0 \\x>2[/tex] Therefore, domain of [tex]f(x)= x^2+3x-10[/tex] here is all values of x > 2 .