Y=√(x+6) - 7
constraint: x+6> or = 0
or x> 0r = -6
RANGE = x∈R/ x > or = 0
The correct answer is the first option
The answer is
A,B
Explanation
Because the square root of A is 4 and the square root of B is 5
The square root of C is 11.1803398875 and the square root of D is 14.6969384567.
Therefore A,B are perfect squares
Answer:
21
Step-by-step explanation:
when p is 4
p² + 5 = 4² + 5
16 + 5 = 21
Answer:
Step-by-step explanation:
Given that acceleration of an object is
![\frac{dv}{dt} =-2v\\\frac{dv}{v} =-2dt\\ln v = -2t+C\\](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D-2v%5C%5C%5Cfrac%7Bdv%7D%7Bv%7D%20%3D-2dt%5C%5Cln%20v%20%3D%20-2t%2BC%5C%5C)
is the solution to the differential equation
Since v(0) =7
we get ln 7 = C
Hence ![lnv = -2t+ln 7\\v=7e^{-2t}](https://tex.z-dn.net/?f=lnv%20%3D%20-2t%2Bln%207%5C%5Cv%3D7e%5E%7B-2t%7D)
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
![C=7/2](https://tex.z-dn.net/?f=C%3D7%2F2)
So solution for distance is
![s(t) =\frac{-7}{2} (e^{-2t}-1)](https://tex.z-dn.net/?f=s%28t%29%20%3D%5Cfrac%7B-7%7D%7B2%7D%20%28e%5E%7B-2t%7D-1%29)