To express the statement "length is 4 times the width plus 4" as a formula, you can define the width as ( w ) and the length as ( l ). The formula can then be written as:
[ l = 4w + 4. ]
This equation indicates that the length (( l )) is equal to four times the width (( w )) plus four.
Length times Width (L x W) is a simple equation for area calculation.
It's width to length ratio is 3 : 8.
To write an expression for the area of a rectangle, use the formula (A = l \times w), where (l) is the length and (w) is the width. You can express this in different ways, such as (A = w \times l) or (A = l^2) if it's a square where (l = w). To find the area, simply substitute the values of length and width into your chosen expression and calculate the result. For example, if (l = 5) and (w = 3), then (A = 5 \times 3 = 15) square units.
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.
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.
To write an expression for the area of a rectangle, use the formula (A = l \times w), where (l) is the length and (w) is the width. You can express this in different ways, such as (A = w \times l) or (A = l^2) if it's a square where (l = w). To find the area, simply substitute the values of length and width into your chosen expression and calculate the result. For example, if (l = 5) and (w = 3), then (A = 5 \times 3 = 15) square units.
Let the length of the rectangle be represented by ( l ). Then, the width can be expressed as ( w = l - 6 ). The area ( A ) of the rectangle is given by the product of its length and width, so ( A = l \times w = l \times (l - 6) ). This simplifies to the quadratic function ( A(l) = l^2 - 6l ).
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.
Let the length of the painting be represented by the variable ( L ). Given that the width is 5 inches shorter than the length, the width can be expressed as ( W = L - 5 ). Thus, the algebraic expressions for the length and width of the painting are ( L ) and ( L - 5 ), respectively.