A teacher decided to bring a jar of 530 pieces of small candy to his 100-student classroom so students could practice estimation
1 answer:
You might be interested in
Recall that variation of parameters is used to solve second-order ODEs of the form
<em>y''(t)</em> + <em>p(t)</em> <em>y'(t)</em> + <em>q(t)</em> <em>y(t)</em> = <em>f(t)</em>
so the first thing you need to do is divide both sides of your equation by <em>t</em> :
<em>y''</em> + (2<em>t</em> - 1)/<em>t</em> <em>y'</em> - 2/<em>t</em> <em>y</em> = 7<em>t</em>
<em />
You're looking for a solution of the form
where
and <em>W</em> denotes the Wronskian determinant.
Compute the Wronskian:
Then
The general solution to the ODE is
which simplifies somewhat to
Answer:
A) a vertical line does not represent a function.
Step-by-step explanation:
For a relation to be a function for each value of there must be only one value of
. In other words a function is one in which each value in the domain set corresponds to only one value in the range set.
Let us check for this condition in the give choices:
A) a vertical line
A vertical line is given as which meas it is parallel to y-axis and has infinite number of
values for a single
value.
So, its Not a function
B)
For the given equation, on plugging in some value will give a single
value.
So, its a Function
C) a horizontal line
A horizontal line is given as which meas it is parallel to x-axis and has infinite number of
values giving a single
value.
So, its a Function
D) {(1, 7), (3,7), (5, 7), (7,7)}
For the given set for different valuesthere is only one
value.
So, its a Function