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What is the quadratic equation and area of a rectangle when its length is 7 cm greater than its width which is 3x cm?
Wiki User
∙ 12y ago
Quadratic equation: 9x2-21x+12.25 = 0
Area: 36.75 square cm
Wiki User
∙ 12y ago
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How do you find the length and width of a rectangle when given the area and perimeter?
By forming a quadratic equation from the information given and then the length and width can be found by solving the equation.
The length of a rectangle is 2 inches greater than the width The are is 80 inches2 Find the length of the rectangle?
Let the length be x+2 and the width be x: (x+2)*x = 80 inches2 x2+2x = 80 Rearrange the equation into a quadratic: x2+2x-80 = 0 Solve by factoring or with the help of the quadratic equation formula: (x+10)(x-8) = 0 So x = -10 or x = 8 it must be the latter because the dimensions can't be negative. Therefore: length = 10 inches and the width = 8 inches
What is the answer of equation that length multiplied by width of a rectangle?
area of the rectangle..
The length of a rectangle is 2 inches greater than the width The area is 80 inches2. Find the length of the rectangle?
Let the length be x+2 and the width be x: (x+2)*x = 80 x2+2x = 80 x2+2x-80 = 0 Solving the above with the quadratic equation formula works out as: x = -10 or x = 8, so it must be the latter because dimensions can't be negative. Therefore: length = 10 inches
How do you find the diagonal of a rectangle when its width is 4.2 cm shorter than its length and has an area of 211.68 cm?
In order to find the diagonal the length and width of the rectangle must be found first so let the length be x and the width be (x-4.2) length*width = area x*(x-4.2) = 211.68 Multiply out the brackets and subtract 211.68 from both sides thus forming a quadratic equation: x2-4.2-211.68 = 0 Solving the above equation by means of the quadratic formula gives a positive value of 16.8 for x Therefore: length = 16.8 cm and width = 12.6 cm Use Pythagoras to find the rectangle's diagonal: 16.82+12.62 = 441 and the square root of this is 21 Therefore the diagonal of the rectangle is 21 cm
How do you find the length and width of a rectangle when given the area and perimeter?
By forming a quadratic equation from the information given and then the length and width can be found by solving the equation.
The length of a rectangle is 2 inches greater than the width The are is 80 inches2 Find the length of the rectangle?
Let the length be x+2 and the width be x: (x+2)*x = 80 inches2 x2+2x = 80 Rearrange the equation into a quadratic: x2+2x-80 = 0 Solve by factoring or with the help of the quadratic equation formula: (x+10)(x-8) = 0 So x = -10 or x = 8 it must be the latter because the dimensions can't be negative. Therefore: length = 10 inches and the width = 8 inches
The length of a rectangle is 4 meters greater than the width. The area of the rectangle is 96 meters square. Find the length and width.?
Let the length be x+4 and the width be x:- length*width = area (x+4)*x = 96 square meters Multiply out the brackets and subtract 96 from both sides thus forming a quadratic equation:- x2+4x-96 = 0 Solving the equation using the quadratic formula gives x a positive value of 8 Therefore: length = 12 meters and width = 8 meters Check: 12*8 = 96 square meters
What is the answer of equation that length multiplied by width of a rectangle?
area of the rectangle..
The length of a rectangle is 2 inches greater than the width The area is 80 inches2. Find the length of the rectangle?
Let the length be x+2 and the width be x: (x+2)*x = 80 x2+2x = 80 x2+2x-80 = 0 Solving the above with the quadratic equation formula works out as: x = -10 or x = 8, so it must be the latter because dimensions can't be negative. Therefore: length = 10 inches
What is the size of the diagonal of a rectangle when its length is 7 cm greater than its width and has an area of 60 square cm?
13 cm Solved with the help of the quadratic formula and Pythagoras' theorem.
How do you find the diagonal of a rectangle when its width is 4.2 cm shorter than its length and has an area of 211.68 cm?
In order to find the diagonal the length and width of the rectangle must be found first so let the length be x and the width be (x-4.2) length*width = area x*(x-4.2) = 211.68 Multiply out the brackets and subtract 211.68 from both sides thus forming a quadratic equation: x2-4.2-211.68 = 0 Solving the above equation by means of the quadratic formula gives a positive value of 16.8 for x Therefore: length = 16.8 cm and width = 12.6 cm Use Pythagoras to find the rectangle's diagonal: 16.82+12.62 = 441 and the square root of this is 21 Therefore the diagonal of the rectangle is 21 cm
Can width of a rectangle be greater than the length?
yes
The area of a rectangle is 45 square feet. The length is 4 feet longer than the width. Which equation could you use to find the length of the rectangle in feet?
Let the length be x+4 and the width be x:- Form a quadratic equation: (x+4)*x = 45 => x^2 +4x -45 = 0 Solving the equation: x has a positive value of 5 Therefore length is 5+4 = 9 feet and width = 5 feet Check: 9 times 5 = 45 square feet
What is the proper equation to calculate the area of a rectangle?
length * width
The length of a rectangle is 2 inches more than its width the area of the rectangle is 15 square inches what are the length and width of the rectangle?
Call the width of the rectangle x. Call the length of the rectangle x + 2. The area of the rectangle would be the length times the width. We know this is 15 square inches. Set up the equation and solve for x. (x +2)x = 15 x2 + 2x = 15 x2 + 2x - 15 = 0 Factor the quadratic equation: (x - 3)(x + 5) = 0 x = 3 or x = -5 Since length can't be negative, discard the answer -5. If x = 3, then the width is 3 and the length is x + 2 or 5. 5 x 3 = 15 which equals the area.
The width of a rectangle is 11 cm mire than the length the area of the rectangle is 1302 cm label which is the length and which is the width show work too please?
W = width of rectangle L = length of rectangle A = area of rectangle W x L = A (L+11) x L = A L squared + 11 L - 1302 = 0 solve for L by quadratic equation or by factoring: (L-31)(L +42) = 0 L = 31 W = L+11 = 42
