Hello,
For a given perimeter (P) there are an infinity of Area (A)
Let's say x the length, and y the wide of the rectangle
P=2(x+y)
A=xy
k=x-y >=0
As (x+y)²-4xy=(x-y)²: A²-4P=k² or P=(A²-k²)/4
In primus, you will find a graph (abacus) giving P for a A and k given.
Negative Area or P are excluded.(just remind the first quadrant, A>=0 and P>=0)
Answer:
To find the mean absolute deviation of the data, start by finding the mean of the data set. Find the sum of the data values, and divide the sum by the number of data values. Find the absolute value of the difference between each data value and the mean: |data value – mean|.
<h2>please make me as brainlylist </h2>
Consider the two functions as
<span>y1(x) =3x^2 - 5x,
y2(x) = 2x^2 - x - c
The higher the value of c, father apart the two equations will be.
They will touch when the difference, i.e. y1(x)-y2(x)=x^2-4*x+c has a discriminant of 0.
This happens when D=((-4)^2-4c)=0, or when c=4.
(a)
So when c=4, the two equations will barely touch, giving a single solution, or coincident roots.
(b)
when c is greater than 4, the two curves are farther apart, thus there will be no (real) solution.
(c)
when c<4, then the two curves will cross at more than one location, giving two distinct solutions.
It will be more obvious if you plot the two curves in a graphics calculator using c=3,4, and 5.
</span>
Answer:
the domain is 5 and the range is 10
Step-by-step explanation: