I'm assuming you mean (a-b)^2 : (a+b)^2
a^2 + b^2 cancel out leaving
-2ab versus +2ab
Therefore (a+b)^2 is +4ab greater.
24 and 28
These are the only two numbers (other than 20 itself) that have a factor of 4. And if you were to find all of their common factors, 4 would be the greatest!
Step-by-step explanation:
y=-1
-1 + 4 = 3
y = x - 6
-1 - 6= -7
Answer:
-7
Answer:
and 
and b=
Step-by-step explanation:
We are given that a quadratic equation

We have to solve the equation by completing square and find the value of b.






and 

And 
Therefore,
and 
and b=
Smaller leg = x
longer leg = y
hypotenuse = z
y=x+7
z=x+8
due to

but x cannot be -3
so smaller leg is 5 inches
then longer leg = 5+7 = 12 inches
and hypotenuse = 5+8 = 13 inches