What is the area of a sector when r=9/2 and 0=5pi/6 radians?
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Its algebra. The original equation is 
To solve for a variable, we reverse the order of operations, beginning with addition/subtraction, and then multiplication/division. To remove a number from one side, we must do the opposite to the other side. In this case, to get rid of the -121 we must add 121 to the -164. This gives us -43. Then, to get the x by itself, we must multiply the other side by 3. -43*3=129
When we are doing the opposite of an operation to the other side, we are really reversing the operation and, to keep both sides equal, we must do whatever we have done to one side to the other side. So when we have -121, we add 121 as it equals 0, therefore it is gone. Since a equation must be balanced, we have to do what we did to the other side (adding 121).
To solve for a variable, we reverse the order of operations, beginning with addition/subtraction, and then multiplication/division. To remove a number from one side, we must do the opposite to the other side. In this case, to get rid of the -121 we must add 121 to the -164. This gives us -43. Then, to get the x by itself, we must multiply the other side by 3. -43*3=129
When we are doing the opposite of an operation to the other side, we are really reversing the operation and, to keep both sides equal, we must do whatever we have done to one side to the other side. So when we have -121, we add 121 as it equals 0, therefore it is gone. Since a equation must be balanced, we have to do what we did to the other side (adding 121).