An equation has infinitely many solutions if it can be manipulated all the way to an identity (i.e. an equality where the right and left hand side are the same). We have:
A) ![12+4x=6x+10-2x \iff 12+4x=10+4x \iff 12=10](https://tex.z-dn.net/?f=12%2B4x%3D6x%2B10-2x%20%5Ciff%2012%2B4x%3D10%2B4x%20%5Ciff%2012%3D10)
which is impossible
B) ![5x+14-4x=23+x-9 \iff x+14 = x+14](https://tex.z-dn.net/?f=5x%2B14-4x%3D23%2Bx-9%20%5Ciff%20x%2B14%20%3D%20x%2B14)
which is an equality
C) ![x+9-0.8x=5.2x+17-8\iff 0.2x+9=5.2x+9 \iff 0.2x=5.2x\iff x=0](https://tex.z-dn.net/?f=x%2B9-0.8x%3D5.2x%2B17-8%5Ciff%200.2x%2B9%3D5.2x%2B9%20%5Ciff%200.2x%3D5.2x%5Ciff%20x%3D0)
which has a unique solution
D) ![4x-2x=20 \iff 2x=20 \iff x=10](https://tex.z-dn.net/?f=4x-2x%3D20%20%5Ciff%202x%3D20%20%5Ciff%20x%3D10)
which has a unique solution
Answer:
10%
Step-by-step explanation:
We have been given that the manager of a store orders model railroad sets that cost $390 and sells them for $429.
We will use markup % formula for our given problem.
![\text{Markup percentage}=\frac{\text{Selling price- Cost}}{\text{Cost}}\times 100](https://tex.z-dn.net/?f=%5Ctext%7BMarkup%20percentage%7D%3D%5Cfrac%7B%5Ctext%7BSelling%20price-%20Cost%7D%7D%7B%5Ctext%7BCost%7D%7D%5Ctimes%20100)
![\text{Markup percentage}=\frac{429-390}{390}\times 100](https://tex.z-dn.net/?f=%5Ctext%7BMarkup%20percentage%7D%3D%5Cfrac%7B429-390%7D%7B390%7D%5Ctimes%20100)
![\text{Markup percentage}=\frac{39}{390}\times 100](https://tex.z-dn.net/?f=%5Ctext%7BMarkup%20percentage%7D%3D%5Cfrac%7B39%7D%7B390%7D%5Ctimes%20100)
![\text{Markup percentage}=\frac{1}{10}\times 100](https://tex.z-dn.net/?f=%5Ctext%7BMarkup%20percentage%7D%3D%5Cfrac%7B1%7D%7B10%7D%5Ctimes%20100)
![\text{Markup percentage}=10](https://tex.z-dn.net/?f=%5Ctext%7BMarkup%20percentage%7D%3D10)
Therefore, markup is 10% of cost.
I believe it should be 22½ or 22 books
You would need to divide 18 and ⅘
18 ÷ ⅘ = 22½
I hope this helps, even though I’m not totally sure :)
72 times 4 = 288
answer- 288