Answer:
A. b(w) = 80w +30
B. input: weeks; output: flowers that bloomed
C. 2830
Step-by-step explanation:
<h3>Part A:</h3>
For f(s) = 2s +30, and s(w) = 40w, the composite function f(s(w)) is ...
b(w) = f(s(w)) = 2(40w) +30
b(w) = 80w +30 . . . . . . blooms over w weeks
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<h3>Part B:</h3>
The input units of f(s) are <em>seeds</em>. The output units are <em>flowers</em>.
The input units of s(w) are <em>weeks</em>. The output units are <em>seeds</em>.
Then the function b(w) above has input units of <em>weeks</em>, and output units of <em>flowers</em> (blooms).
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<h3>Part C:</h3>
For 35 weeks, the number of flowers that bloomed is ...
b(35) = 80(35) +30 = 2830 . . . . flowers bloomed over 35 weeks
Answer: Region A
Step-by-step explanation:
The pressure is lower at region A because it shows less particles at the higher the chart goes.
Answer: Work done by the particle is 78 N.
Step-by-step explanation:
If C is the path the particle follows, then work done is 
.
According to the question the force 
and all four points are in the plane 
.
therefore, if S is the at surface with boundary C, so that S is the portion of the plane over the rectangle D=[0,2]\times [0,4].
Now,
By the Stoke's theorem:
![\int_{C}^{}F.dx=\iint_{s}^{}curlF.dS\\\\=\iint_{D}^{}[-8y(0)-2z\left ( \frac{1}{4} \right )+5y]dA\\\\=\iint_{D}^{}\left ( -\frac{1}{2}z+5y \right )dA](https://tex.z-dn.net/?f=%5Cint_%7BC%7D%5E%7B%7DF.dx%3D%5Ciint_%7Bs%7D%5E%7B%7DcurlF.dS%5C%5C%5C%5C%3D%5Ciint_%7BD%7D%5E%7B%7D%5B-8y%280%29-2z%5Cleft%20%28%20%5Cfrac%7B1%7D%7B4%7D%20%5Cright%20%29%2B5y%5DdA%5C%5C%5C%5C%3D%5Ciint_%7BD%7D%5E%7B%7D%5Cleft%20%28%20-%5Cfrac%7B1%7D%7B2%7Dz%2B5y%20%5Cright%20%29dA)
![\int_{0}^{2}\left [ \frac{39}{16}y^2 \right ]_{0}^{4}dx\\\\=\int_{0}^{2}39\, \, dx\\\\=39\times 2\\\\=78 \, N](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7B2%7D%5Cleft%20%5B%20%5Cfrac%7B39%7D%7B16%7Dy%5E2%20%5Cright%20%5D_%7B0%7D%5E%7B4%7Ddx%5C%5C%5C%5C%3D%5Cint_%7B0%7D%5E%7B2%7D39%5C%2C%20%5C%2C%20%20dx%5C%5C%5C%5C%3D39%5Ctimes%202%5C%5C%5C%5C%3D78%20%5C%2C%20N)
