Answer:
The number of horses that can eat 4 stacks of hay in 8 days = 56 horses
Step-by-step explanation:
The given parameters are;
The time it takes 16 horses to eat 5 stacks = 35 days
Therefore;
The time it takes 16 horses to eat 5/5 stacks (1 stack) = 35 days/5 = 7 days
The time it takes 16 horses to eat 1 stack of hay = 7 days
The time it takes 16 horses/16 to eat 1 stack of hay = 7 days × 16 = 112 days
Therefore;
The time it takes 1 horse to eat 1 stack of hay = 112 days
The time it takes 1 horse to eat 4 × 1 stack of hay = 112 days × 4 = 448 days
The time it takes 1 horse to eat 4 stacks of hay = 448 days
Therefore, given that (448 days)/(8 days/horse) = 56 horse, we have;
The number of horses that will eat 4 stacks of hay in 8 days = 56 horses.
Answer:
<h2>384 yd²</h2>
Step-by-step explanation:
The formula of a surface area of a cube:
We have <em>a = 8 yd</em>.
Substitute:
Answer:
-7
Step-by-step explanation:
y−x=9 ⇒ y=x+9
10+2x=−2y
5+x= -y ====== replacing y with x+9
5+x= -x-9
2x= -14
x= -7
Answer: 36.65 inches
Step-by-step explanation:
The length of arc with a central angle x and radius r in a circle is given by :-
Given : Radius of a circle = 60 inches
Central angle =
Now, the of length of arc is given by :-
Hence, the length of arc = 36.65 inches.
Answer:
Remember that:
Speed = distance/time.
Then we can calculate the average speed in any segment,
Let's make a model where the average speed at t = t0 can be calculated as:
AS(t0) = (y(b) - y(a))/(b - a)
Where b is the next value of t0, and a is the previous value of t0. This is because t0 is the middle point in this segment.
Then:
if t0 = 100s
AS(100s) = (400ft - 0ft)/(200s - 0s) = 2ft/s
if t0 = 200s
AS(200s) = (1360ft - 50ft)/(300s - 100s) = 6.55 ft/s
if t0 = 300s
AS(300s) = (3200ft - 400ft)/(400s - 200s) = 14ft/s
if t0 = 400s
AS(400s) = (6250s - 1360s)/(500s - 300s) = 24.45 ft/s
So for the given options, t = 400s is the one where the velocity seems to be the biggest.
And this has a lot of sense, because while the distance between the values of time is constant (is always 100 seconds) we can see that the difference between consecutive values of y(t) is increasing.
Then we can conclude that the rocket is accelerating upwards, then as larger is the value of t, bigger will be the average velocity at that point.
