What is the expression for the area of a square with side 'x'?I have a side of square that denoted by x.
I have also an other square that his side is (x + m).
I want to show that the area of (x+m)-square bigger If m > 0 from the x-square.
How can I prove it?
Substitution is one way to prove it.I have a side of square that denoted by x.
I have also an other square that his side is (x + m).
I want to show that the area of (x+m)-square bigger If m > 0 from the x-square.
How can I prove it?
x^2 and (x+m)^2=x^2 + 2xm + m^2.What is the expression for the area of a square with side 'x'?
What is the expression for the area of a square with side 'x+m'?
You are not thinking!x^2 and (x+m)^2=x^2 + 2xm + m^2.
I see that by subscription, As Jeff wrote.You are not thinking!
How do you prove "some number" is bigger than another number?
As m > 0, (x+m)-x is positive. Which means (x+m) > xWhat am I asking for: I have a paper with 2 squares. One is bigger than the other. How can I denote the notion x and x+m (m > 0) to each square?
What level of work are you doing here? You posted this under Geometry so I would presume that if m > 0 we may simply state that x + m > x. If you need more:O.K.
I look at my question and I need to go deeper. (I didn't notice that I ask the question that I want to ask).
I have 2 squares one has side that is x and the another has side is that x+m.
If m > 0 then how I justified that the bigger square is with side x+m? How can I prove it? What assumptions I need to write so I can say for sure that x is of the smaller square?
[Sorry about that I don't ask the correct question (that I want to ask).]
What is:x^2 and (x+m)^2=x^2 + 2xm + m^2.