The width is 18 cm and the length is 20 cm. The new dimensions are 15 by 24.
The formula used is if w=width, w (w + 2) = (w - 3) (w + 6).
Assuming that the length of a rectangle is greater than its breadth, any value L such that 15 ≤ L < 30 cm will do. Then breadth, B = 30 - L cm.
any no. between 0&1 when raised to some power will obviously decreased.
No, but language implies it.
Yes.
The area of a rectangle does not provide enough information. Let L be any number greater than sqrt(60) = approx 7.75 and let B = 60/L. Then a rectangle with length L units and Breadth B units will have an area of L*B = L*(60/L) = 60 square units. Since L can be any number greater than 7.75, there are infiitely many choices for L and so infinitely many possible rectangles.
Usually, yes.
Assuming that the length of a rectangle is greater than its breadth, any value L such that 15 ≤ L < 30 cm will do. Then breadth, B = 30 - L cm.
Perimeter = 2*(Length + Breadth) So, 1570 = 2*(168 + Breadth) 785 = 168 + Breadth Breadth = 617 metres Which is rather an unusual result because normally the length is greater than the breadth.
any no. between 0&1 when raised to some power will obviously decreased.
twelve -------------------- The original rectangle is 12 feet in length and 5 feet in width. The length (12) is 7 feet greater than the width (5 feet). Were the length decreased by 3 (12 - 3 = 9) and the width increased by 2 (5 + 2 = 7), the perimeter would be 32 feet (9 + 7 + 9 + 7 = 32).
No, but language implies it.
Yes.
10x + 5 > 12x - 2 7 > 2x 7/2 > x
Of course, a rectangle can have a greater perimeter and a greater area. Simply double all the sides: the perimeter is doubled and the area is quadrupled - both bigger than they were.
The area of a rectangle does not provide enough information. Let L be any number greater than sqrt(60) = approx 7.75 and let B = 60/L. Then a rectangle with length L units and Breadth B units will have an area of L*B = L*(60/L) = 60 square units. Since L can be any number greater than 7.75, there are infiitely many choices for L and so infinitely many possible rectangles.
The way the rectangle is usually defined, a rectangle has to be positive, so any number greater than zero will do. Note that the set of numbers greater than zero has no minimum.
yes it can; a rectangle 5 by 2 has perimeter 14 and area 10 for example; a rectangle 10 by 2 has perimeter 24 and area 20, both greater.