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.
Step-by-step 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;
![\begin{gathered} A_1=l_1\ast w_1=6(1)/(4)\ast2(2)/(3)=(25)/(4)\ast(8)/(3)=(25)/(1)\ast(2)/(3)=(50)/(3) \\ \therefore A_1=(50)/(3)=16(2)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/uijwp4l82j1ruqnre9gkgormg6k9w6gg0e.png)
Therefore, the area of the 1st sign is 50/3 ft^2
Let the area of the second sign be A2;
![\begin{gathered} A_2=l_{2_{}}\ast w_2 \\ =4(3)/(4)\ast3=(19)/(4)\ast3=(57)/(4) \\ \therefore A_2=(57)/(4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/k37mxbnj9ysv7v6nw4gbm8vhychgz73enz.png)
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;
![\begin{gathered} A_1=(50\ast4)/(3\ast4)=(200)/(12) \\ A_2=(57\ast3)/(4\ast3)=(171)/(12) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4af7ou59nq5qiiu6dbwebl006oifpdyk4a.png)
Since the numerator of A1 is greater than that of A2, it means that A1 is the largest.