Answer:
Option C is correct.
Explanation:
The following research laboratory publishes their new work concerning articles of information allocated to the better standards container classification of information. The community manager needs to ensure all of those posts appear across the Tips and Methods subject field for Community users.
Thus, set up an automatic subject assignment and map the list of Tips and Techniques to the Container good practices group.
The answer to that is a Pixel
Applying potential difference to a conductor, by potential force, free electrons gain energy and move from low to high potential. Thus, electrons move from one atom to another.
Answer:An initial condition is an extra bit of information about a differential equation that tells you the value of the function at a particular point. Differential equations with initial conditions are commonly called initial value problems.
The video above uses the example
{
d
y
d
x
=
cos
(
x
)
y
(
0
)
=
−
1
to illustrate a simple initial value problem. Solving the differential equation without the initial condition gives you
y
=
sin
(
x
)
+
C
.
Once you get the general solution, you can use the initial value to find a particular solution which satisfies the problem. In this case, plugging in
0
for
x
and
−
1
for
y
gives us
−
1
=
C
, meaning that the particular solution must be
y
=
sin
(
x
)
−
1
.
So the general way to solve initial value problems is: - First, find the general solution while ignoring the initial condition. - Then, use the initial condition to plug in values and find a particular solution.
Two additional things to keep in mind: First, the initial value doesn't necessarily have to just be
y
-values. Higher-order equations might have an initial value for both
y
and
y
′
, for example.
Second, an initial value problem doesn't always have a unique solution. It's possible for an initial value problem to have multiple solutions, or even no solution at all.
Explanation:
Answer:
B,C,E your welcome
Explanation: i just took the test
