A rectangle has a perimeter of 80 cm and its length is 1 cm more than twice its width. Find the dimensions of a rectangle given that its perimeter is 80 cm and its length is 1 cm more than twice its width. Set up your solution using the variables L for the length, W for the width, and P for the perimeter. Part a: Using the definition of the perimeter, write an equation for P in terms of L and W. Part b: Using the relationship given in the problem statement, write an equation for L in terms of W. Solve the equations from parts a and b. Part c: The length is ? Cm. Part d: The width is? Cm.

(a) P = 6W + 2

(b) L = 2W + 1

(c) Width = 13cm

Length = 27cm

Step-by-step explanation:

The formula for perimeter of a rectangle is 2(length + width). Since the length is 1 cm more than twice its width, then the length will be:

L = (2 × W) + 1

(b) L = 2W + 1

Therefore, P = 2(L + W)

P = 2( 2W + 1) + 2W

P = 4W + 2 + 2W

(a) P = 6W + 2

Since perimeter is given as 80cm. Therefore,

P = 6W + 2

6W + 2 = 80

6W = 80 - 2

6W = 78

W = 78/6

W = 13

Width is 13cm

Length = 2W + 1

Length = 2(13) + 1

Length = 27cm

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.