A rounded rectangle is not a function at all.
Area of a rectangle is length x width. It isn't clear what the width is in this case - or how you could solve for it.
Area of rectangle: (x+5)(3x-7) = 3x2+8x-35
crossed rectangle is not a rectangle, rectangle have to have 90 degree angles.
Every rectangle must have corners otherwise it would not be a rectangle!Every rectangle must have corners otherwise it would not be a rectangle!Every rectangle must have corners otherwise it would not be a rectangle!Every rectangle must have corners otherwise it would not be a rectangle!
A rounded rectangle is not a function at all.
The size of the circle is a function of the height of the rectangle.
The average value of a function is the equivalent area of the function expressed like a rectangle .- Thus If you have an area A from limits a and b , the equivalent is : (b-a) H ( rectangle area ) = INT f(x) dx , then average H is H= ( INT f(x) dx) /(b-a)
There will be a function in it like this: double RectangleArea (double a, double b) { return a*b; }
Area of a rectangle is length x width. It isn't clear what the width is in this case - or how you could solve for it.
Area of rectangle: (x+5)(3x-7) = 3x2+8x-35
The area of a rectangle with a width of x units and a length of (x + 3) units
Suppose the length and width of the rectangle are L and W metres respectively.Then the perimeter, P = 20 m implies that2(L + W ) = 20 => L + W = 10 or W = 10 - L.Then Area = L * W = L * (10 - L) sq metres.
If the perimiter is 20 and one side is [[length]] then the other side is (10 - [[length]]). So the area is: [[length]] x (10 - [[length]]) square metres.
crossed rectangle is not a rectangle, rectangle have to have 90 degree angles.
This question has no unique answer. A (3 x 2) rectangle has a perimeter = 10, its area = 6 A (4 x 1) rectangle also has a perimeter = 10, but its area = 4 A (4.5 x 0.5) rectangle also has a perimeter = 10, but its area = 2.25. The greatest possible area for a rectangle with perimeter=10 occurs if the rectangle is a square, with all sides = 2.5. Then the area = 6.25. You can keep the same perimeter = 10 and make the area anything you want between zero and 6.25, by picking different lengths and widths, just as long as (length+width)=5.
Rectangle area = (rectangle width) x (rectangle height)