Question 3: Find d^2ydx^2 for y=4x^4−3x^3−6x^2+x A d^2ydx^2=16x^3−9x^2−12x+1

B d^2ydx^2=16x^2−9x−12

C d^2ydx^2=48x^2−18x−12

D d^2ydx^2=8x^2−6x−8

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

50x5=$250

step-by-step explanation:

80 the answer is 80

step-by-step explanation:

answer: 8d

step-by-step explanation:

45+5= 50+5=55+5=60