Given the function f(x)=x^2+5x+5 , to get the zeros we solve using quadratic formula;
x=[-b(+or-) sqrt(b^2-4ac)]/(2a)
from our function;
a=1,b=5, c=5
thus,
x=[-5(+or-)sqrt(5^2-4*1*5)]/(2*1)
x=[-5(+or-)sqrt(25-20)]/2
x=[-5+/-sqrt(5)/2
The answer is option C
Answer:
u are fricked my dude
Step-by-step explanation:
Answer:
No, (3, 9) is not a solution of the system.
Step-by-step explanation:
y=3x
2x-y=6
-----------
3(3)=9
2(3)-9=6-9=-3, not 6
Vertices (3,0),(-3,0) co-vertices (0,-5),(0,5)
transverse axis (line passing vertices) is on(or parallel to) x-axis then formula is
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
..notice.. x^2 is on positive / y^2 is on negative
center (h,k) is midway between vertices = (0,0)
we have h = k = 0 and now formula is
x^2/a^2 - y^2/b^2 = 1
a is the distance from a vertex to center = 3
b is the distance from a co-vertex to center = 5
the formula is
x^2/3^2 - y^2/5^2 = 1 ... answer is the 1st