I've attached a plot of the intersection (highlighted in red) between the parabolic cylinder (orange) and the hyperbolic paraboloid (blue).
The arc length can be computed with a line integral, but first we'll need a parameterization for

. This is easy enough to do. First fix any one variable. For convenience, choose

.
Now,

, and

. The intersection is thus parameterized by the vector-valued function

where

. The arc length is computed with the integral

Some rewriting:

Complete the square to get

So in the integral, you can substitute

to get

Next substitute

, so that the integral becomes

This is a fairly standard integral (it even has its own Wiki page, if you're not familiar with the derivation):

So the arc length is
It must have the [( )]
[(4/y) - z]
Answer:
Apple
Orange
Step-by-step explanation:
Apple has the greater median with a median value of 150 which is more than that of orange at 130
The sample with the greater variability is the orange sample as the interquartile range value of 11 is greater Than 8 for Apple.
Yes it is possible, because of the greater variation on weight shown by the sample of oranges, then it could be possible.
<h3>
Answer: k = 3</h3>
===========================================================
Explanation:
- A = first term = k+2
- B = second term = 4k-6
- C = third term = 3k-2
To go from the first term to the second term, we add on some common difference d.
So,
B = A+d
B = (k+2)+d
4k-6 = k+2+d
4k-6-k-2 = d
d = 3k-8
---------------
Similarly, to go from the second term to the third term, we also add on d
C = B+d
C = (4k-6)+d
C = (4k-6)+(3k-8)
C = 7k-14
3k-2 = 7k-14
---------------
Let's solve for k
3k-2 = 7k-14
-2+14 = 7k-3k
12 = 4k
4k = 12
k = 12/4
k = 3 is the final answer
---------------
If k = 3, then we have these three terms:
- A = k+2 = 3+2 = 5
- B = 4k-6 = 4(3)-6 = 6
- C = 3k-2 = 3(3)-2 = 7
The arithmetic progression (AP) is 5, 6, 7. The common difference is d = 1.
Note how d = 3k-8 = 3(3)-8 = 1