Answer:
[tex]0\leq x\leq 15[/tex]
Step-by-step explanation:
Given is the graph with equation
[tex]y=-2x^2+30x+200[/tex]
This should be atleast 200 implies[tex]-2x^2+30x+200\geq 200\\x(-2x+30)\geq 0[/tex]
This is the product of two numbers hence would be positive only if either both are positive or both are negative
Case I: Both positive .
[tex]x\geq 0 and -2x+30\geq 0\\0\leq x\leq 15[/tex]
Case II: Both negative
Then we get
[tex]x\leq 0 and -2x+30\leq 0\\\\x\leq 0 and x\geq 15[/tex]
This is inconsistent as a value cannot be less than 0 and greater than 15
So solution is
[tex]0\leq x\leq 15[/tex]
0 is less than or equal to x is than or equal to 15, B is the correct answer!
