U varies inversely which means p³ will come as a denominator of the constant k.
Therefore, The equation formed will be:

Move all terms that don't contain y to the right side and solve.
y = - x/3 + 14/3
Hope this helps! :)
Answer: 113.04
Step-by-step explanation: The equation for the area of a circle is pi r squared so you have to find the radius first. To find the radius, find the diameter first. A way to do this which was easiest for me was dividing 37.68 by pi (3.14), which is 12. Divide the diameter by 2 to find the radius, 6.
Now use the equation from the beginning, pi r squared. So plug in pi as 3.14, then the radius as 6. It is important to always square the radius first, which would be 36. Now multiply 36 by 3.14, and there should be your answer.
Set the following: Length = L Width = W
The length is twice the width: L = 2*W or: W = L/2
Perimeter: P = 60ft = 2*L + 2*W
Substituting L for W in the perimeter equation gives:
P = 60ft = 2*L + 2*(L/2) = 2*L + L = 3L
L = 60ft/3 = 20ft W=L/2=20ft/2=10ft
Verify result: Perimeter= P = 60ft = 2L + 2W = 2*20ft +2*10ft = 40ft+20ft = 60ft
The area of the rectangular is: Area = A = Length*Width = L*W = 20ft*10ft = 200 (ft)^2
-Rosie
Start by writing the system down, I will use
to represent 

Substitute the fact that
into the first equation to get,

Simplify into a quadratic form (
),

Now you can use Vieta's rule which states that any quadratic equation can be written in the following form,

which then must factor into

And the solutions will be
.
Clearly for small coefficients like ours
, this is very easy to figure out. To get 5 and 6 we simply say that
.
This fits the definition as
and
.
So as mentioned, solutions will equal to
but these are just x-values in the solution pairs of a form
.
To get y-values we must substitute 3 for x in the original equation and then also 2 for x in the original equation. Luckily we already know that substituting either of the two numbers yields a zero.
So the solution pairs are
and
.
Hope this helps :)