If we multiply by negative number on both sides of inequality n sign of inequality will change i.e. if x>7 n (-1) x < (-1)7 => -x<-7.

Since it has equality sign along with inequality sign so it is a slack inequality not strict inequality.

3x-6 ≥ 0 => x ≥ 2.
(0,0) does not satisfy te equation so region is right side of x=2 with complete line x=2 due to presence of equality sign along with inequality sign.

10x?50
Dividing by 10 on both sides, x ? (50/10) => x?5
Since x is a whole number so x = 0,1,2,3,4,5.

Amount left is at least 10 rupees i.e. amount left ≥ 10.
x-40 ≥ 10 => x ≥ 50.

Let shortest side be x. Then longest side = 2x.
Third side = 2x-4.
Given, perimeter of triangle is at least 61 cm
=>x+2x+2x-4 ≥ 61 => 5x≥65 = x≥13.
Minimum length of shortest side is 13 cm.

x>7 and 7>4 => x>7>4 => x>4.
If x>7 n x cannot be less than 6.
If x=11 n x>7 and x>9.
If x=11 n x>7 and x<14.

F = 9/5 C + 32°
C=(F-32°)*5/9
77° ≤ F ≤ 86°
=> 77°-32° ≤ F-32° ≤ 86° -32°
=> 45° ≤ F-32° ≤ 54°
=>45o*5/9 ≤ (F-32°) *5/9 ≤ 54°*5/9
=>25° ≤ C ≤ 30°.

x-1>-x+7
=>2x>8 => x>4.

Since region involves 1^st quadrant so x≥0, y≥0.
Two points on line are (0,6) and (3,0).
(y-6)/(0-6) = (x-0)/(3-0)
=>(y-6)/(-6) = x/3
=>y-6=-2x => 2x+y=6
2x+y≤6 since (0,0) should also satisfy.
So, 2x+y≤6, x≥0, y≥0.