Which function has real zeros at x = 3 and x = 7? f(x) = x2 + 4x – 21 f(x) = x2 – 4x – 21 f(x) = x2 – 10x + 21 f(x) = x2 – 10x – 21

Option 3rd is correct

Step-by-step explanation:

As per the statement

We have a real zeros at x = 3 and x = 7

Since, the graph have zeros at x = 3 and x = 7 then;

By zero product property: if ab= 0 then a = 0 or b =0

⇒ and x -7 = 0

⇒

Let f(x) be the function has a real zeros at x = 3 and x = 7 then;

⇒

Using distributive property,

we have;

Combine like terms;

Therefore, the function has real zeros at x = 3 and x = 7 is,

the answer is b.)

< 1=< 2

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.