From the edge of a 486-foot cliff, peyton shot an arrow over the ocean with an initial upward velocity of 90-feet per second

9 seconds

Step-by-step explanation:

The complete question is

The altitude of an object, d, can be modeled using the equation below:

d=-16t^2 +vt+h

from the edge of a 486 foot cliff, Peyton shot an arrow over the ocean with an initial upward velocity of 90 feet per second. In how many seconds will the arrow reach the water below?

Let

d ----> the altitude of an object in feet

t ---> the time in seconds

v ---> initial velocity in ft per second

h ---> initial height of an object in feet

we have

we  know that

When the arrow reach the water the value of d is equal to zero

we have

substitute the values and solve for t

Multiply by -1 both sides

The formula to solve a quadratic equation of the form is equal to

in this problem we have

substitute in the formula

the solution is t=9 sec

see the attached figure to better understand the problem

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

the reporter's right

step-by-step explanation:

1/4 = .25 / 25%