Answer:
, which corresponds to answer D.
Step-by-step explanation:
Recall that the maximum or minimum of a parabola is always located at its vertex. Notice that all quadratic functions listed are given in what is called "vertex" form, since they explicitly show the vertex coordinates in their formulation:
where 
, and 
stand for the x and y coordinates respectively of the vertex.
Examining all four options, we notice that in all four cases listed, the is correct (equal to positive 9), so we proceed to examine what the 
expression should look like for 
:
Notice that when replace with "-3", we end up with a sign change:
Therefore we find that the last two options can be candidates. Then we recall that the question also states that this should be a minimum. Then, for a parabola to have a minimum, its branches must go upwards, and this corresponds to a case in which the factor 
(leading coefficient) multiplying (x-x_v)^2 is positive. So in looking for such condition, we find that the very last option is the only one that verifies such.
Then, is the option we select.
Answer:
The answer is y=3 or y=6
Step-by-step explanation:
So basically the formula is:
3/5 = 12/n
so then we get 3n = 60
n = 20
so there are 20 picks in all
The equation is actually
circle area = PI * radius^2
radius = square root (area / PI)
The picture in the attached figure
we know that
RS=ST----------------> because is a square base
so
QS is the diagonal of the square base
applying the Pythagoras Theorem
QS²=RS²+ST²------> 6²+6²-----> QS²=72------> QS=√72----> QS=6√2 units
the answer isQS=6√2 units
