The answer is 7.57 of what they make for each year
Given that a flyer begins from earth and thrown up with a velocity of 30 ft/sec vertically.
u = initial velocity = 30 : a = -g = 32 ft/sec^2
s(0) =initial height =4 ft.
We have the equation
where u = initial velocity : v= final velocity : s = distance travelled and a = acceleration. Here final velocity is found out as follows
Substitute to get
Since v cannot be negative, the flyer’s center of gravity cannot ever reach 20 feet.
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To reach 25 ft, put s = 25
and v must be atleast 0
Then we have
Hence if thrown with initial velocity of 40 ft /sec, it will reach a height of 25 ft.
The differential equation
![M(x,y) \, dx + N(x,y) \, dy = 0](https://tex.z-dn.net/?f=M%28x%2Cy%29%20%5C%2C%20dx%20%2B%20N%28x%2Cy%29%20%5C%2C%20dy%20%3D%200)
is considered exact if
(where subscripts denote partial derivatives). If it is exact, then its general solution is an implicit function
such that
and
.
We have
![M = \tan(x) - \sin(x) \sin(y) \implies M_y = -\sin(x) \cos(y)](https://tex.z-dn.net/?f=M%20%3D%20%5Ctan%28x%29%20-%20%5Csin%28x%29%20%5Csin%28y%29%20%5Cimplies%20M_y%20%3D%20-%5Csin%28x%29%20%5Ccos%28y%29)
![N = \cos(x) \cos(y) \implies N_x = -\sin(x) \cos(y)](https://tex.z-dn.net/?f=N%20%3D%20%5Ccos%28x%29%20%5Ccos%28y%29%20%5Cimplies%20N_x%20%3D%20-%5Csin%28x%29%20%5Ccos%28y%29)
and
, so the equation is indeed exact.
Now, the solution
satisfies
![f_x = \tan(x) - \sin(x) \sin(y)](https://tex.z-dn.net/?f=f_x%20%3D%20%5Ctan%28x%29%20-%20%5Csin%28x%29%20%5Csin%28y%29)
Integrating with respect to
, we get
![\displaystyle \int f_x \, dx = \int (\tan(x) - \sin(x) \sin(y)) \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20f_x%20%5C%2C%20dx%20%3D%20%5Cint%20%28%5Ctan%28x%29%20-%20%5Csin%28x%29%20%5Csin%28y%29%29%20%5C%2C%20dx)
![\implies f(x,y) = -\ln|\cos(x)| + \cos(x) \sin(y) + g(y)](https://tex.z-dn.net/?f=%5Cimplies%20f%28x%2Cy%29%20%3D%20-%5Cln%7C%5Ccos%28x%29%7C%20%2B%20%5Ccos%28x%29%20%5Csin%28y%29%20%2B%20g%28y%29)
and differentiating with respect to
, we get
![f_y = \cos(x) \cos(y) = \cos(x) \cos(y) + \dfrac{dg}{dy}](https://tex.z-dn.net/?f=f_y%20%3D%20%5Ccos%28x%29%20%5Ccos%28y%29%20%3D%20%5Ccos%28x%29%20%5Ccos%28y%29%20%2B%20%5Cdfrac%7Bdg%7D%7Bdy%7D)
![\implies \dfrac{dg}{dy} = 0 \implies g(y) = C](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac%7Bdg%7D%7Bdy%7D%20%3D%200%20%5Cimplies%20g%28y%29%20%3D%20C)
Then the general solution to the exact equation is
![f(x,y) = \boxed{-\ln|\cos(x)| + \cos(x) \sin(y) = C}](https://tex.z-dn.net/?f=f%28x%2Cy%29%20%3D%20%5Cboxed%7B-%5Cln%7C%5Ccos%28x%29%7C%20%2B%20%5Ccos%28x%29%20%5Csin%28y%29%20%3D%20C%7D)
Answer:
(a) 8
(b) 16
(c) 24
(d) The ratios are the same
Step-by-step explanation:
(a) For inputs -5 and -4, the difference of outputs is ...
-3 -(-11) = 8
__
(b) For inputs 0 and 2, the difference of outputs is ...
45 -29 = 16
__
(c) For inputs -3 and 0, the difference of outputs is ...
29 -5 = 24
__
(d) The ratios of output difference over input difference are ...
8/1 = 16/2 = 24/3 = 8 . . . . . they are all the same
This is yet another confirmation that the slope of a line is the same everywhere, and over any interval.