Answer:
Consider f: N → N defined by f(0)=0 and f(n)=n-1 for all n>0.
Step-by-step explanation:
First we will prove that f is surjective. Let y∈N be any natural number. Define x as the number x=y+1. Then x∈N, and f(x)=x-1=(y+1)-1=y. We conclude that f is surjective.
However, f is not injective. Take x1=0 and x2=1. Then x1≠x2 but f(x1)=0 and f(x2)=x2-1=1-1=0. We have shown that there are two natural numbers x1,x2 such that x1≠x2 but f(x1)=f(x2), that is, f is not injective.
Note:
If 0∉N in your definition of natural numbers, the same reasoning works with the function f: N → N defined by f(1)=1 and f(n)=n-1 for all n>1. The only difference is that you consider x1=1, x2=2 for the injectivity.
Step-by-step explanation:
a) gradient= y2-y1/x2-x1
gradient of AB:
gradient=-4-1/2-3
=-5/-1
=5
gradient of BC:
=-5+4/7-2
=-1/5
b) as the priduct of both gradient is -1 ,both lines are perpendicular.
c) as we have seen lines are perpendicular so it is a right angled triangle.
Answer:
=26x^3−12x^2+5x+7
Step-by-step explanation:
2x^3−3x+11−(3x^2+1)(4−8x)
Distribute:
=2x^3+−3x+11+24x^3+−12x^2+8x+−4
Combine Like Terms:
=2x^3+−3x+11+24x^3+−12x2+8x+−4
=(2x^3+24x^3)+(−12x^2)+(−3x+8x)+(11+−4)
=26x^3+−12x^2+5x+7