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.
When the radius is increased, the period of rotation will increase. This is because a larger radius means the object has to travel a greater distance in the same amount of time, leading to a longer period of rotation.
any no. between 0&1 when raised to some power will obviously decreased.
When the effort force is decreased, the mechanical advantage must be increased in order to maintain the same level of output force. This can be achieved by either adjusting the length of the lever or using different mechanical systems that provide a greater advantage.
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.
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.
It is not possible to answer the question.Let L be any number greater than 24 units and let B = 24 - L. Then the perimeter of a rectangle with length L and breadth B is2*(L + B) = 2*(L + 24 - L) = 2*24 = 48.The choice of L is arbitrary and so there are infinitely many possible solutions.