Pretend you and Will owe a friend instead of the bank, pull out your wallets and pay your debts. Who has to take out more money from their wallet? Of course, Will has to take out $9.25 more than Chaz.
Lets say we have a rectangle. if the ratio is 7:5, we can put the width as 7x, and the length as 5x (or vice versa, it doesn't really matter.) If the perimeter is solved by 2(h+w), we have 2(7x+5x)=96, or 2(7x)+2(5x)=96, which is C
Answer:

Step-by-step explanation:
So the thing about adding of integers is that is both nos. are positive then you can simply add it and same with negative too.But with negative and positive we should subtract but that is for later.Luckily we have 2 positive integers so we do it like this:
+6 is the same as 6 and +2 is the same as 2 so we...
rule:add the numbers and add the sign.

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.
1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.
According to this theorem,

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

Take positive value x. You get

2. According to the previous theorem,

Then

Answer: 
This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

This means that you cannot find solutions of this equation. Then CD≠2 cm.