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Length times Width (L x W) is a simple equation for area calculation.
It's width to length ratio is 3 : 8.
I assume you mean that the length is less than four times the width. Here is the outline. First assume, for simplicity, that the length is EQUAL to 4 times the width. Write two equatios for that - one for the perimeter, one for the area. Solve it, and find the corresponding area. That's basically the minimum area. At the other extreme, make the length equal to the width, and solve again. That would be the maximum area.
The area is the length times the width. That's the function. If you want to write a function in a computer language, you need two parameters. Just return the product of the two parameters. Example in Java: double rectangle_area(double length, double width) { return length * width; } I didn't test this, but that's the basic idea.
The measurements for a box is written thus: length, width and depth.
Length X Width X Height
It doesn't really matter, but most people usually write length x width.
Length times Width (L x W) is a simple equation for area calculation.
It's width to length ratio is 3 : 8.
61:71 1:71/61
When calculating volume, write down the formula you are using, the measurements or dimensions of the object or space, and show your calculations step by step. Make sure to include the unit of measurement in your answer.
Let's denote the width of the rectangle as ( w ) and the length as ( 3w ). The perimeter of a rectangle is given by the formula ( P = 2l + 2w ), where ( l ) is the length and ( w ) is the width. Given that the perimeter is 64 cm, we can write the equation ( 64 = 2(3w) + 2w ). Solving this equation, we find that the width is 8 cm and the length is 24 cm.
I assume you mean that the length is less than four times the width. Here is the outline. First assume, for simplicity, that the length is EQUAL to 4 times the width. Write two equatios for that - one for the perimeter, one for the area. Solve it, and find the corresponding area. That's basically the minimum area. At the other extreme, make the length equal to the width, and solve again. That would be the maximum area.
The area is the length times the width. That's the function. If you want to write a function in a computer language, you need two parameters. Just return the product of the two parameters. Example in Java: double rectangle_area(double length, double width) { return length * width; } I didn't test this, but that's the basic idea.
The measurements for a box is written thus: length, width and depth.
Having the length x meters and width 8 meters, then the area is 8x square meters.
It is given that the length is twice the width plus 3 yards; this can be written as a formula: Length = 2 × Width + 3 To calculate the area of a rectangle the formula is: Area_rectangle = Length × Width Substitute for Length in the above formula by the given expression involving Width → Area = Length × Width = (2 × Width + 3) × Width = 2 × Width² + 3 × Width But the area is given as 119 sq yd. If I now write w for Width, this gives: 2w² + 3w = 119 sq yd → 2w² +3w - 119 = 0 Which is a quadratic which can be solved either by using the formula or by factorising: It does factorise: the factor pairs of 119 are 1 × 119 and 7 × 17. One bracket will have (2w ...) and the other will have (w...). As the 119 is negative, there are effectively four pairs of numbers that could be fitted into the brackets: {-1, 119}; {-7, 17}; {7, -17}; {1, 119}. There is one pair which will fit so that when multiplied out gives the quadratic. A quadratic will always have two solutions; as one of the numbers above is negative and one positive, the two solutions will comprise of one positive and one negative value for w (Width); as a length must be positive, the negative solution cannot work and the positive solution is the required Width, from which the Length can be calculated (as above).