Check if the equation is exact, which happens for ODEs of the form

if
.
We have


so the ODE is not quite exact, but we can find an integrating factor
so that

<em>is</em> exact, which would require


Notice that

is independent of <em>x</em>, and dividing this by
gives an expression independent of <em>y</em>. If we assume
is a function of <em>x</em> alone, then
, and the partial differential equation above gives

which is separable and we can solve for
easily.




So, multiply the original ODE by <em>x</em> on both sides:

Now


so the modified ODE is exact.
Now we look for a solution of the form
, with differential

The solution <em>F</em> satisfies


Integrating both sides of the first equation with respect to <em>x</em> gives

Differentiating both sides with respect to <em>y</em> gives


So the solution to the ODE is


2. W<5
4. D<6 (or equal to)
The equation of the line that is parallel to the given line and passes through the point (-2,2) is:

<h3>What is a linear function?</h3>
A linear function is modeled by:
y = mx + b
In which:
- m is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.
- b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.
When two lines are parallel, they have the same slope. In this problem, the given line passes through (-5,-4) and (0,-3), hence the slope is:
m = (-3 - (-4))/(0 - (-5)) = 1/5.
Hence the equation is:

When x = -2, y = 2, then:



Hence:

More can be learned about linear equations at brainly.com/question/24808124
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Use the pythagorean theorem to solve this: a^2 + b^2 = c^2.
You already have a and b which are 3 and 4. Square them up and you will get 9 + 16 = c^2.
Add.
25 = c^2
Lastly, square root both sides.
5 = c
The length of AB is 5.
Answer:

Step-by-step explanation:
Given
, start by squaring both sides to work towards isolating
:

Recall
and
:

Isolate the radical:

Square both sides:

Expand using FOIL and
:

Move everything to one side to get a quadratic:

Solving using the quadratic formula:
A quadratic in
has real solutions
. In
, assign values:

Solving yields:

Only
works when plugged in the original equation. Therefore,
is extraneous and the only solution is 