Let:
T = Total money they have
Dc = Dog collar
x = Number of bags of food
![\begin{gathered} 2x+Dc\le T \\ 2x+4.5\le30 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%202x%2BDc%5Cle%20T%20%5C%5C%202x%2B4.5%5Cle30%20%5Cend%7Bgathered%7D)
Let's solve the inequality:
![\begin{gathered} 2x+4.5\le30 \\ \text{subtract 4.5 from both sides:} \\ 2x+4.5-4.5\le30-4.5 \\ 2x\le25.5 \\ x\le\frac{25.5}{2} \\ x=12.75\approx12 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%202x%2B4.5%5Cle30%20%5C%5C%20%5Ctext%7Bsubtract%204.5%20from%20both%20sides%3A%7D%20%5C%5C%202x%2B4.5-4.5%5Cle30-4.5%20%5C%5C%202x%5Cle25.5%20%5C%5C%20x%5Cle%5Cfrac%7B25.5%7D%7B2%7D%20%5C%5C%20x%3D12.75%5Capprox12%20%5Cend%7Bgathered%7D)
This means that they can buy 12 bags of food at most.
Since they plan to buy 2 bags of food:
![\begin{gathered} 2(7)+4.5\le30 \\ 14+4.5\le30 \\ 18.5\le30 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%202%287%29%2B4.5%5Cle30%20%5C%5C%2014%2B4.5%5Cle30%20%5C%5C%2018.5%5Cle30%20%5Cend%7Bgathered%7D)
This means they can do it and have $ 11.5 to spare
The power of 10 is -5
Move the decimal point to the right until the first significant number and in this problem it is 2. It took 5 places to the right until going past the 2 and the number would be negative because going right past the decimal point will make any power negative.
What is there to solve
be more specific
Find the greatest common factor and then divide each number by it.
in this case the gcf for 75 and 100 is 25.
75 divided by 25 = 3
100 divided by 25 = 4
so 75/100 simplified = 3/4