Sorry. It's not clear. Plz tell me if there is 12 3/8in?
Observe the graph below. This graph represents the scenario.
The question is ill formated, the complete question is
In a simulation, a moving object accelerates from rest to 4 meters per second in 2 seconds. For the following three seconds, it increases linearly until it reaches a speed of 10 meters per second. Following three seconds at that speed (acceleration = 0), the item progressively decelerates until it comes to rest two seconds later. Draw the graph of this scenario for 10 seconds?
I'll describe how the graph may show.
It will move diagonally upward from time 0 to 2 seconds until it reaches the y axis at a speed of 4 m/s.
Then, from 2 to 5, the position will move up diagonally until it reaches the y axis at a speed of 10 m/s.
The next 5 to 8 seconds will be horizontal.
After that, it will descend diagonally.
Observe the graph below. This graph represents the scenario.
Learn more about Acceleration here-
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To do these problems, plug in a couple of values into the equations, and see the general shape of the graph. To ensure you were right, check them on an online graphing calculator. I highly recommend Desmos Graphing Calculator. Cheers!
We use P = i•e^rt for exponential population growth, where P = end population, i = initial population, r = rate, and t = time
P = 2•i = 2•15 = 30, so 30 = 15 [e^(r•1)],
or 30/15 = 2 = e^(r)
ln 2 = ln (e^r)
.693 = r•(ln e), ln e = 1, so r = .693
Now that we have our doubling rate of .693, we can use that r and our t as the 12th hour is t=11, because there are 11 more hours at the end of that first hour
So our initial population is again 15, and P = i•e^rt
P = 15•e^(.693×11) = 15•e^(7.624)
P = 15•2046.94 = 30,704
Im not sure but you cam try adding 15+9 which is 24