Ok, so user says that it should be solve for vertex not vertex form
(x,y)
to find the vertex of
y=ax^2+bx+c
the x value of the vertex is -b/2a
the y value is found by plugging in the x value for the vertex back into the original equation and evaluating
y=-2x^2-12x-28
a=-2
b=-12
xvalue of vertex is -(-12)/(2*-2)=12/-4=-3
x value of vertex is -3
plug backin for x
y=-2x^2-12x-28
y=-2(-3)^2-12(-3)-28
y=-2(9)+36-28
y=-18+8
y=-10
yvalue is -10
x value is -3
vertex is (-3,-10)
I think the answer would be B
Answer:
sqrt of 2/2
Step-by-step explanation:
Answer:
The first choice is correct.
Step-by-step explanation:
First work out the common ratio r:-
a2 = a1r and a5 = a1 r^4
So r^3 = a1r^4 / a1r = 256/ 108
r = cube root (256/-108) = 4/3
So the explicit formula is 108(4/3)^(n-1)
Answer:
whole number
Step-by-step explanation:
its not a decimal and its not a fraction
