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Sagot :
To solve for the value of x in the length of a rectangle:
[tex]\begin{gathered} \text{Length = (2x+1)cm} \\ \text{width = 12cm} \\ \text{Area is not less than 108cm}^2 \end{gathered}[/tex]Area of a rectangle = length x width , If the area of a rectangle is less than 108 thus the value is greater than 108cm²
[tex]\begin{gathered} \text{Area of a rectangle = length x width } \\ A=\text{ L X W} \\ A>\text{ L x W} \end{gathered}[/tex][tex]\begin{gathered} (2x+1\text{ x 12)>108} \\ 24x+12>108 \\ 24x>108-12 \\ 24x>96 \\ x>\frac{96}{24} \\ x>4 \end{gathered}[/tex]Hence the correct answer for the value of x > 4
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