Decimals are simply a way to express numbers. There is nothing to solve!
You take the formula for a rectangle, replace what you know, and solve for the remaining variable.
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Add together the lengths of each of its sides.
Make sure you know the length and width of the rectangle in the SAME UNITS ... both in inches, or both in meters etc. Then just multiply the two numbers ... Length x Width. The answer is the area of the rectangle.
In a rectangle: P = 2(L + W) and A = LW P = 2(L + W) (replace P with 20) 20 = 2(L + W) (solve for L; divide by 2 to both sides) 10 = L + W (subtract W to both sides ) 10 - W = L A = LW (substitute 10 - W for L) A = (10 - W)W A = 10W - W2
There is no "height" of a rectangle, unless it's a rectangular prism. Do you mean the length? If you have the area of the rectangle, the equation should be:A= L x WPlug in the area and the length and solve for the width, or plug in the area the width of the rectangle, and solve for the length.
It really depends on the specific function, and what you want to solve for.
Decimals are simply a way to express numbers. There is nothing to solve!
You take the formula for a rectangle, replace what you know, and solve for the remaining variable.
You can certainly express the radius as a function of its area, yes. If the area is known, you can solve the formula for the area of the circle to uniquely get the radius. (The quadratic equation has two solutions; you will of course choose the positive solution for the radius.)
More information is needed to solve this problem
Add together the lengths of each of its sides.
For a rectangle, it is 2(l + w) l = length w = width
Write an equation for the perimeter, and solve it. Remember that the perimeter is the sum of all four sides.
Make sure you know the length and width of the rectangle in the SAME UNITS ... both in inches, or both in meters etc. Then just multiply the two numbers ... Length x Width. The answer is the area of the rectangle.
Write two simultaneous equations and solve them. One for the perimeter, one for the area.