Step-by-step explanation:
Comparision: Let say you buy chocolate that cost $1. And there was luckily a coupon, for 10% off.
That means after applying the coupon, you will only pay 90% of the original price. Which is
[tex]0.9(1) = 0.9[/tex]
which equals 90 cents.
So let go back to this problem and apply similar thinking.
The price of the original root is g, so for the original price column, check the 3rd cell, g.
Use our comparision problem, by applying the coupon to the chocolate, you will save
[tex]1 - 0.9 = 0.1[/tex]
which is 1/10 of the original price.
In this equation, we can represent this as 0.1g.
So for the second column, shade in 5th cell, 0.1g.
The price would be
[tex](0.9)1 = 0.9[/tex]
So our equation for this equation will be
0.9g.
So for third column, shade in 1st cell, 0.9g.
