The width of rectangular field is x metres. the length of the field is 30m longer than the width. the perimeter of the field is less than 500m. the area of the field is greater than 4000m^2by writing suitable inequalities, find the possible values of x

Possible values of x are   50 < x < 110.

Step-by-step explanation:

Consider the perimeter of the field:

2x + 2(x + 30) < 500

2x + 2x + 60 < 500

4x < 440

x < 110.

Consider the area of the field:

x(x + 30 ) > 4000

x^2 + 30x - 4000 > 0

(x - 50)(x + 80) > 0

The critical values are x = -80 and 50.

As x is a width it must be positive so x > 50.

tigers

step-by-step explanation:

cougars 12/10 (12-10 = 2)

bobcats 20/10 (20-10 =10)

bulldogs 8/5 (8-5 =3)

tigers 3/5 (3-5 =-2)

for finding the midpoint of any given 2 pair of points simply add the respective x coordinates with other x coordinate and the y coordinate with other y coordinate and divide them by 2

in your question it will be :

(3+1)/2 for the x coordinate of mid point i.e 2

[8+(-4)]/2 for the y coordinate of midpoint i.e 2

so coordinates of mid point of points (3,8) and (1,-4) will be (2,2)