Aball is hurled in the air such that its horizontal (x) and vertical (y) shifts are given by the functions x(t) = 20t and y(t) = 40t − 5t2, where x and y are in meters and the time t is in seconds. what is the ratio of y to x at time t = 5 seconds? 0.25 0.50 0.75 0.85

x(t) = 20t

y(t) = 40t - 5t^2

Since we are only interested in comparing the two at time t = 5 seconds, we plug in 5 everywhere we see the variable t and then compare x and y

x(5) = 20(5) becomes x(5) = 100

y(5) = 40(5) - 5(5)^2 becomes y(5) = 200 - 125 and then y(5) = 75

The ratio of y to x can be expressed as: y/x, so we can say the ratio is equal to 75/100 or 0.75

0.75

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

i have no clue at all sorry

step-by-step explanation: