This equation has only one solution
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P.S. Hello from Russia
Answer:
Step-by-step explanation:
Given that acceleration of an object is
is the solution to the differential equation
Since v(0) =7
we get ln 7 = C
Hence 
since velocity is rate of change of distance s we have
![v=\frac{ds}{dt} =7e^{-2t}\\s= [tex]s(t) =\frac{-7}{2} (e^{-2t})+C)[](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bds%7D%7Bdt%7D%20%3D7e%5E%7B-2t%7D%5C%5Cs%3D%20%5Btex%5Ds%28t%29%20%3D%5Cfrac%7B-7%7D%7B2%7D%20%28e%5E%7B-2t%7D%29%2BC%29%5B)
substitute t=0 and s=0
So solution for distance is
I suspect you may have slipped up as you copied the question. Exactly as you wrote it, it has a single zero, at x= 2 .
Answer:
AECG
Step-by-step explanation:
1
sqrt(49) = 7
sqrt(a^2) = a
sqrt(b^2) = b
For every two variables you can take one out from under the root sign and thorough the other one away.
Answer: E
2
sqrt(36) = 6
sqrt(a^2) = a See comment for 1.
b must be left where it is. There is only 1 of them.
6asqrt(b)
Answer: A
3. sqrt(25) = 5
sqrt(b^2) = b
a must be left alone. There's only 1 of them.
5b sqrt(a)
answer: C
4
sqrt(81 a b)
sqrt(81) = 9
The variables must be left alone. There's only1 of them
9 sqrt(ab)
Answer G
