Step-by-step explanation:
1 .f(x)=-x+1
f(-1)= (-1)-(-1)+1
f(-1)= 1+1+1
= 3
3.f(x)=-x+1
f( 1)=(1)-1+1
f( 1)=1-1+1
=1
5. f(x)=-x+1
f(3)=-3+1
= 9-3+1
=7
2. g(x) = 5 - 3x
g( -8)=5-3(-8)
g( -8)=5+24
=29
4.g(x) = 5 - 3x
g(5)=5-3(5)
g(5)=5-15
=-10
6.g(x) = 5 - 3x
g(-3)=5-3(-3)
g(-3)=5+9
=14
9514 1404 393
Answer:
about 44.5 feet
Step-by-step explanation:
We can write relations for the height of the rail as a function of initial length and expanded length, but the solution cannot be found algebraically. A graphical solution or iterative solution is possible.
Referring to the figure in the second attachment, we can write a relation between the angle value α and the height of the circular arc as ...
h = c·tan(α) . . . . . . where c = half the initial rail length
Then the length of the expanded rail is ...
s = r(2α) = (c/sin(2α)(2α) . . . . . . where s = half the expanded rail length
Rearranging this last equation, we have ...
sin(2α)/(2α) = c/s
It is this equation that must be solved iteratively. We find the solution to be ...
α ≈ 0.0168538794049 radians
So, the height of the circular arc is ...
h = 2640.5·tan(0.0168538794049) ≈ 44.4984550191 . . . feet
The rail will bow upward by about 44.5 feet.
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<em>Additional comments</em>
Note that s and c in the diagram are half the lengths of the arc and the chord, respectively. The ratio of half-lengths is the same as the ratio of full lengths: c/s = 2640/2640.5 = 5280/5281.
We don't know the precise shape the arc will take, but we suspect is is not a circular arc. It seems likely to be a catenary, or something similar.
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We used Newton's method iteration to refine the estimate of the angle from that shown on the graph. The iterator used is x' = x -f(x)/f'(x), where x' is the next guess based on the previous guess of x. Only a few iterations are required obtain an angle value to full calculator precision.
Answer:
A) -9/2
B) 9/4
C) -9/2, same as A)
Step-by-step explanation:
We are given that . We use the properties of integrals to write the new integrals in terms of I.
A) . We have used that ∫cf dx=c∫f dx.
B) . Here we used that reversing the limits of integration changes the sign of the integral.
C) It's the same integral in A)