Answer:
I look down on copyright. To take someone else's work and disguising it as your own seems like a form of theft.
Answer:
- The graph of the function is attached below.
- The x-intercepts will be: (2, 0), (-2, 0)
- The y-intercept will be: (-20, 0)
Explanation:
Given the function
As we know that the x-intercept(s) can be obtained by setting the value y=0
so
switching sides
Add 20 to both sides
Dividing both sides by 5
so the x-intercepts will be: (2, 0), (-2, 0)
we also know that the y-intercept(s) can obtained by setting the value x=0
so
so the y-intercept will be: (-20, 0)
From the attached figure, all the intercepts are labeled.
Answer:
I think it is the last one. Or first. Try first though.
Explanation:
Have a Great Day.
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:
