Given, to find A we have to :
From the question, the following equation can be formed :
Therefore, the formula for finding A = m² + y - 7
<h2><u>Solution (2)</u> :</h2>The value of A we can derive from the formula is :
Value of m = 3 (given)
Which means :
Thus, the value of A = 2+y
Therefore, the value of A = <u>2+y</u>
Part (i)
We have 60 students total, and 5 didn't like any of the two subjects, so that must mean 60-5 = 55 students liked at least one subject.
<h3>Answer: 55</h3>=========================================================
Part (ii)
We have 15 who like math only, 20 who like science only, and 55 who like either (or both). Let x be the number of people who like both classes.
We can then say
15+20+x = 55
x+35 = 55
x = 55-35
x = 20
This means 20 people liked both subjects
<h3>Answer: 20</h3>=========================================================
Part (iii)
There are 15 people who like math only, and 20 who like both. Therefore, there are 15+20 = 35 people who like math (and some of these people also like science)
<h3>Answer: 35</h3>=========================================================
Part (iv)
We'll follow the same idea as the previous part. There are 20 people who like science only and 20 who like both subjects. That yields 40 people total who like science (and some of these people also like math).
<h3>Answer: 40</h3>=========================================================
Part (v)
We'll draw a rectangle to represent the entire group of 60 students. This is considered the universal set. Inside the rectangle will be two overlapping circles to represent math (M) and science (S).
We'll have 15 go in circle M, but outside circle S to represent the 15 people who like math only. Then we have 20 go in circle S but outside circle M to show the 20 people who like science only. We have another copy of 20 go in the overlapped region between the circles. This is the 20 people who like both classes. And finally, we have 5 go outside both circles, but inside the rectangle. These are the 5 people who don't like either subject.
Note how all of the values in the diagram add up to 60
15+20+20+5 = 60
This helps confirm we have the correct values.
<h3>Answer: See the venn diagram below</h3>
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.
_____
<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.
