Show that 21^n, n €N can never ends with 0,2,4,6 and 8​

Step-by-step explanation:

21 has no 2 in it's prime factorization. So it can never end in 0, 2 , 4 , 6 , 8.

 OR

Odd number raise to any number is always odd, Which does not include 2, 4, 6 , 8.

So does not end with 0, 2, 4 , 6 , 8.

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

f(2x−x2)=−3x2+6x−1

step-by-step explanation:

set up the composite function and evaluate.

f(2x−x2)=−3x2+6x−1