B. and D. represent functions
Answer:Part a) 
Part b) When Jenny divides the square root of her favorite positive integer by
, the result is an integer.
Step-by-step explanation:
Let
x-------> the favorite positive integer
Part a)
1) For 
-----> the product results in an integer
thus
The number
could potentially be Jenny's favorite positive integer
2) For 
-----> the product results in an integer
thus
The number
could potentially be Jenny's favorite positive integer
3) For 
-----> the product results in an integer
thus
The number
could potentially be Jenny's favorite positive integer
Part B)
1) For 
-----> the outcome is an integer
2) For 
-----> the outcome is an integer
3) For 
-----> the outcome is an integer
Therefore
When Jenny divides the square root of her favorite positive integer by
, she obtains an integer as a result.
Answer:
(C) 10% to 70%(
Step-by-step explanation:
Given that at least 40% of the students are learning German, the upper limit of those who might be enrolled in English but not in German is 60%. However, since a minimum of 70% study English, it leads to the conclusion that at least 10% of students must be taking both German and English.
If we consider that at least 30% of students are learning Italian, and assuming that no student is studying all three languages simultaneously, then there is a maximum of 70% of students who could potentially be registered in both English and German.
This means the possible percentage for students enrolled in both English and German ranges from 10% to 70%
y2 = C1xe^(4x) Step-by-step explanation: Knowing that y1 = e^(4x) satisfies the differential equation y'' - 8y' + 16y = 0, we need to derive the second solution y2 using the reduction of order technique. Let y2 = uy1. Since y2 is a solution to the differential equation, it holds that y2'' - 8y2' + 16y2 = 0. By substituting for y2, its derivatives become y2 = ue^(4x), y2' = u'e^(4x) + 4ue^(4x), and y2'' = u''e^(4x) + 8u'e^(4x) + 16ue^(4x). Plugging these into the differential equation gives us u''e^(4x) = 0. Let w = u', so w' = u''. This results in w' e^(4x) = 0, leading to w' = 0. Integrating gives w = C1. Since w = u', this implies u' = C1, and integrating once more results in u = C1x. Therefore, y2 = ue^(4x) becomes y2 = C1xe^(4x), which is the second solution.