The height, h(t), in feet of an object thrown into the air with an initial upward velocity of 63 feet per second is given by the formula h(t) = -16t2 + 63t, where t is the time in seconds. what is the height, in feet, of the object after 3 seconds?

The height of the object after 3 seconds is 45 feets.

Step-by-step explanation:

The height of an object thrown into the air with an initial upward velocity of 63 feet per second is given as

Where, h(t) is height of the object in feet and t is the time in seconds.

We have to find the height of the object after 3 seconds. So, substitute t=3.

Therefore height of the object after 3 seconds is 45 feets.

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!

c

step-by-step explanation: