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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?
Quadratic equation: 9x2-21x+12.25 = 0
Area: 36.75 square cm
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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 base equation for calculating the area of a rectangle?
The base equation for calculating the area of a rectangle is length multiplied by width.
The area of a rectangle is 243 cm2 The length is three times greater than the width What are the dimensions?
Let the length be 3x and the width be x: 3x*x = 243 3x2 = 243 Divide both sides by 3: x2 = 81 Square root both sides: x = 9 Therefore: length = 27 cm and width = 9 cm
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.
If the length of a rectangle is eight feet more than its width and the area of the rectangle is 345 what is the length and width of the rectangle?
Let's denote the width of the rectangle as x feet. Given that the length is eight feet more than the width, we can represent the length as x + 8 feet. The formula for the area of a rectangle is length multiplied by width, so we have the equation (x + 8) * x = 345. By solving this quadratic equation, we find that the width is 15 feet and the length is 23 feet.
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
What is the proper equation to calculate the area of a rectangle?
length * width
