What is 40<= 5/9 (F-32)<= 50 ?

The solution to 40<= 5/9 (F-32)<= 50 is 104<= F<= 122

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40<= 5/9 (F-32)<= 50

40 \leq \frac{5}{9} (F - 32) \leq 50

104 \leq F \leq 122

Interval Notation

[104, 122]

Solution steps

40 \leq \frac{5}{9} (F - 32) \leq 50

a \leq u \leq b

then

a \leq u and u \leq b

40 \leq \frac{5}{9} (F - 32) and \frac{5}{9} (F - 32) \leq 50

40 \leq \frac{5}{9} (F - 32) : F \geq 104

\frac{5}{9} (F - 32) \leq 50 : F \leq 122

Combine the intervals

F \geq 104 and F \leq 122

Merge Overlapping Intervals

104 \leq F \leq 122

n^{2} - 3 n - 40

\frac{5}{2} + 1

\frac{5}{x - 3} > \frac{3}{x + 1}

\frac{2}{3} + \frac{2}{7}

6 x^{4} - 10 x^{3} - 6 x^{2}

What is 40<= 5/9 (F-32)<= 50 ?
The solution to 40<= 5/9 (F-32)<= 50 is 104<= F<= 122