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Sagot :
Answer:
Area of 1st sign is 50/3 ft^2.
Area of 2nd sign is 57/4 ft^2.
Area of the 1st sign is larger than that of the 2nd sign.
Explanation:
To find the area of the 1st sign, we have to multiply the length and width of the sign given;
Let the area of the 1st sign be A1;
[tex]\begin{gathered} A_1=l_1\ast w_1=6\frac{1}{4}\ast2\frac{2}{3}=\frac{25}{4}\ast\frac{8}{3}=\frac{25}{1}\ast\frac{2}{3}=\frac{50}{3} \\ \therefore A_1=\frac{50}{3}=16\frac{2}{3} \end{gathered}[/tex]Therefore, the area of the 1st sign is 50/3 ft^2
Let the area of the second sign be A2;
[tex]\begin{gathered} A_2=l_{2_{}}\ast w_2 \\ =4\frac{3}{4}\ast3=\frac{19}{4}\ast3=\frac{57}{4} \\ \therefore A_2=\frac{57}{4} \end{gathered}[/tex]Therefore, the area of the 2nd sign is 57/4 ft^2.
To determine which fraction is greater, we have to make them both have a common denominator by multiplying the numerator and denominator of A1 by 4 and that of A2 by 3;
[tex]\begin{gathered} A_1=\frac{50\ast4}{3\ast4}=\frac{200}{12} \\ A_2=\frac{57\ast3}{4\ast3}=\frac{171}{12} \end{gathered}[/tex]Since the numerator of A1 is greater than that of A2, it means that A1 is the largest.
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