Answer:
25 ft^2
Step-by-step explanation:
In direct variation, if y varies directly with x, then the equation has the form
y = kx,
where k is the constant of proportionality. y is proportional to x.
Let's call the area y and the distance x.
Here, the area varies with the square of the distance, so the equation has the form
y = kx^2
Here, y is proportional to the square of x.
We can find the value of k by using the given information.
y = kx^2
When x = 20 ft, y = 16 ft^2.
16 = k(20^2)
k = 16/400
k = 1/25
The equation of the relation is:
y = (1/25)x^2
Now we use the equation we found to answer the question.
What is y (the area) when x (the distance) is 25 ft?
y = (1/25)x^2
y = (1/25)(25^2)
y = 25
Answer: 25 ft^2
Answer:
good to know. but what's the question?
Im not sure what this is, or how to answer it
We have that
coordinates are
<span>A(12,7), B(12,−3), C(−2,−3), and D(−2,7)
</span>
i use the formula of midpoint <span>between A and B
</span>Xm=(x1+x2)/2 Ym=(y1+y2)/2
A(12,7) B(12,−3)
Xm=(12+12)/2=12
Ym=(7+(-3))/2=2
the point (12,2) <span>is halfway between A and B</span>
It looks like the differential equation is

Check for exactness:

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

*is* exact. If this modified DE is exact, then

We have

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

The modified DE,

is now exact:

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

Integrate both sides of the first condition with respect to <em>x</em> :

Differentiate both sides of this with respect to <em>y</em> :

Then the general solution to the DE is
