So the given series is "16, 06, 68, 88, __"
Count all the cyclical opening in each of these numbers. For example in 16, there is a one cyclical loop present in it(the one in 6), similarly in 06 it is two(one in zero and one in 6), going ahead, in 68 it is 3(one in 6 and two in 8).
From here on things become simple: hence, the cyclical figures in these equations written down becomes 1,2,3,4,_,3.
Let's now try solving the above sequence, going by the logical reasoning the only number that can fill in the gap should be 4.
Point R and point T have same x coordinate so their distance is by y axis.
Ty-Ry=2.4-1.3=1.1
So the distance between two points is 1.1
S(p) = 400 - 4p + 0.00002p^4
D(p) = 2800 - 0.0012p^3
S(p) = D(p)
400 - 4p + 0.00002p^4 = 2800 - 0.0012p^3
0.00002p^4 + 0.0012p^3 - 4p - 2400 = 0
p = $96.24
-2y^2 + 6y + 2 = 0
a = -2, b = 6, c = 2
x = (- b + - sqrt(b^2-4ac))/(2a)
x=(-6+-sqrt(36-4*-2*2))/-4
x=(-6+-sqrt(36+16))/-4
x=(-6+-sqrt(52))/-4
x = (-6 +- 2sqrt13)/-4
x = (3 + - sqrt13)/2
