Answer:
8
Step-by-step explanation:
Since the forklift has a maximum capacity of three boxes per trip, simply divide the total number of boxes (23 boxes) by three and round up to the nearest whole unit to find the number of trips required:

Casey had to visit the dock 8 times. He moved three boxes in seven trips and two boxes in one trip.
Hello,
I am going to remember:
y'+3y=0==>y=C*e^(-3t)
y'=C'*e^(-3t)-3C*e^(-3t)
y'+3y=C'*e^(-3t)-3Ce^(-3t)+3C*e^(-3t)=C'*e^(-3t) = t+e^(-2t)
==>C'=(t+e^(-2t))/e^(-3t)=t*e^(3t)+e^t
==>C=e^t+t*e^(3t) /3-e^(3t)/9
==>y= (e^t+t*e^(3t)/3-e^(3t)/9)*e^(-3t)+D
==>y=e^(-2t)+t/3-1/9+D
==>y=e^(-2t)+t/3+k
To solve this problem, you simply have to do base 1 * height 1 = base 2 * height 2. Plugging our numbers in gives us 24 * 4 = 5 * height 2. Solving it out, we get 19.2 as our missing height.
0.625 = 0.625 / 1
Numerator = 0.625 × 10 × 10 × 10 = 625
Denominator = 1 × 10 × 10 × 10 = 1000
Numerator / Denominator = 625 / 1000
Our simplifed fraction is:
= 625/1000
= 5/8