Answer:
g(f(x))= (x+3)(x+3)
Step-by-step explanation:
f(x) =x² + 6x +7
g(x)= x + 2
1st Step: Substitute the x in g(x) =x + 2 by the value of f(x)
g(f(x))= (x² + 6x +7) + 2
g(f(x))= x² + 6x + 9
2nd Step: Simplify
g(f(x))= x² + 6x + 9
g(f(x))= (x+3)(x+3)
Answer:
i think the answer is 5
Step-by-step explanation:
if it greater than or equal to its 1 or higher but, its not 10 because it has to be less than 10 5 falls right in the middle of greater that or equal to 1 but less than 10
Let <em>f(x)</em> = <em>x</em>³ + <em>x</em> - 5. <em>f(x)</em> is a polynomial so it's continuous everywhere on its domain (all real numbers). Since
<em>f</em> (1) = 1³ + 1 - 5 = -3 < 0
and
<em>f</em> (2) = 2³ + 2 - 5 = 5 > 0
it follows by the intermediate value theorem that there at least one number <em>x</em> = <em>c</em> between 1 and 2 for which <em>f(c)</em> = 0.
Answer:
B cause it righrhjeueudjss
