Prove that for a triple element x, y, z the distributive law is always satisfied if one of the elements equals 0 or 1, or if two
1 answer:
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Answer:
C. x³+10x²−5x+5
E. 7x⁵+4x²
F. x+8
Step-by-step explanation:
The graph they described should look something like the one i drew below.
So all you have to do is look at the number with the highest exponent.
If the exponent is an odd number and the x is positive then that function will have an end behavior like the picture i posted.
x³ is good
-x³ is bad because of the negative sign
x² is bad because exponent is even
x⁷ is good
123x¹ is good
