"working backwards..." - ???
Is that like the converse?
If x^3 + 2x^2 = x^2(x + 2) < 0, then x + 2 < 0 ... i.e. x < -2
Not sure how that relates to 2x + 5 < 11
x < 3
Hello, didn't know exactly where to post this. Any guidance would be much appreciated. Thanks.
Let a,b, and c be integers and x,y and z be real numbers. Use the technique of working backward from the desired conclusion to prove that
a). [(x+y)/2] ≥(greater than or equal to) (sqrt)√(x*y)
b.) If x^3 + 2*x^2 < 0 , then 2*x + 5 < 11
c). If an isoceles triangle has sides of length x,y and z, where x=y and z=(sqrt)√(2*x*y), then it is right triangle