Given that, 30 < C < 35 ......(i)
and F = \(\frac { 9 }{ 5 } \) C + 32 \(\Rightarrow \)  F - 32 = \(\frac { 9 }{ 5 } \) C \(\Rightarrow \) C = \(\frac { 5}{ 9 } \)(F - 32)
On putting the value of C inequality (i), we get
30 < \(\frac { 5}{ 9 } \)(F - 32)< 35 \(\Rightarrow \) 30\(\times \frac { 9 }{ 5 } \)<\(\frac { 5}{ 9 } \)(F - 32 ) \(\times \) \(\frac { 9 }{ 5 } \)< 35 \(\times \) \(\frac { 9 }{ 5 } \) [multiplying by \(\frac { 9 }{ 5 } \) on each term]
\(\Rightarrow \) 6 \(\times \)9\(\times \)9 \(\Rightarrow \)  54 < F - 32 < 63
\(\Rightarrow \) 54 + 32  [adding 32 in each term]
\(\Rightarrow \) 86 < F < 95
Thus, the required range of temperature is between 86^oF and 95^oF.
His temperature is not normal. Normal temperature of body is 98 ^o6 F. So, he need medical attention.

Let the length of the shortest side of the triangle be x cm.
Then length of the longest side =3x cm.
Thus the length of the third side =(3x−2) cm.
Since the perimeter of the triangle is at least 61 cm,
x+3x+(3x−2)  ≥ 61
⇒7x−2≥61
⇒7x≥61+2
⇒7x≥63⇒x≥9 
Thus the minimum length of the shortest side is 9 cm.

Let at s km below the Earth's surface, the temperature is between 155^oC and 205^oC.
\(\therefore \)  155 < T(s) < 205
\(\Rightarrow \) 155 < 30 + 25(s - 3)< 205
Ans. 8< s < 10

Use \(|x|\ge a\Longrightarrow x\ge a\) or \(x\le -a\)
(\(-\infty\) \(\frac { -3 }{ 2 } \)] \(\cup \) [3, \(\infty \))

We have, 3x - 5 < x + 7
\(\Rightarrow \) 3x -5 + 5 < x + 7 + 5  [adding 5 on both sides]
\(\Rightarrow \) 3x < x + 12
\(\Rightarrow \) 3x -x < x + 12 - x [Subtracting x from both sides]
\(\Rightarrow \) 2x < 12
\(\Rightarrow \) \(\frac { 2x }{ 2 } \) < \(\frac { 12 }{ 2 } \)
\(\Rightarrow \) x < 6
Now if  x is a whole number, then the solution set {0, 1, 2, 3, 4, 5}

{-1, 0, 1, 2, 3,...}

(11, 13), (13, 15), (15, 17)

Let x litres ofwater be added to 1125 litres of 45% acid solution.
Then total quantity of mixture = (1125 + x) litres
\(\frac { 45 }{ 100 } \times 1125+0\times \frac { x }{ 100 } >\frac { 25 }{ 100 } \times \left( 1125+x \right) \) and \(\frac { 45 }{ 100 } \times 1125+0\times \frac { x }{ 100 } <\frac { 30 }{ 100 } \times \left( 1125+x \right) \)
Combining the above inequations, we get
\(\frac { 25 }{ 100 } \times 100\le \frac { 2025\times 100 }{ 4(1125+x) } \le \frac { 30 }{ 100 } \times 100\)
\(\Rightarrow\) \(25\le \frac { 50625 }{ 1125+x } \le 30\)
\(\Rightarrow\) \(25\le \frac { 50625 }{ 1125+x } \) and \(\frac { 50625 }{ 1125+x } \le 30\)
\(\Rightarrow\) 28125 + 25x \(\le\) 50625 and 50625 \(\le\)33750 + 30x
\(\Rightarrow\) 25x \(\le\) 22500 and 30x \(\ge\) 1687.5
\(\Rightarrow\) x \(\le\) 900 and x \(\ge\) 562.5
\(\Rightarrow\) 562.5 \(\le\) x \(\le\) 900

The given inequality 5x +4y \(\le\) 20
draw the graph of the line 5x + 4y = 20

table of values satisfying the equation 5x +4y = 20

Putting (0,0) in the given inequation, we have 5 x 0 +4 x 0 \(\le\) 20 \(\Rightarrow\) 0 \(\le\) 20, which is true
\(\therefore\) Half plane of 5x +4y \(\le\) 20 is towards origin
Also the given inequality ix x \(\ge\) 1
Draw the graph of the line x = 1
Putting (0,0) in the given inequation, we have 0 \(\ge\) 1 which is false
\(\therefore\) Half plane of x \(\ge\) 1 is away from origin
the given inequality is y \(\ge\) 2
Draw the graph of the line y = 2
Putting (0,0) in the given inequation, we have 0 \(\ge\)2, which is false
\(\therefore\) half plane y \(\ge\) 2 is away from origin