Use pascal's triangle to expand the following binomial expression
1.(2k-1/3)⁶​

9514 1404 393

  64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729

Step-by-step explanation:

The row of Pascal's triangle we need for a 6th power expansion is ...

  1, 6, 15, 20, 15, 6, 1

These are the coefficients of the products (a^(n-k))(b^k) in the expansion of (a+b)^n as k ranges from 0 to n.

Your expansion is ...

  1(2k)^6(-1/3)^0 +6(2k)^5(-1/3)^1 +15(2k)^4(-1/3)^2 +20(2k)^3(-1/3)^3 +...

     15(2k)^2(-1/3)^4 +6(2k)^1(-1/3)^5 +1(2k)^0(-1/3)^6

  = 64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729

answer:  

step-by-step explanation:

odd numbers are only divisible by 1 and itself.

even numbers are divisible by more than 1 and itself.

for example 6 is even because it is divisible by more than 1 and its self. 1 x 6= 6 ; 2 x 3= 6

and 7 is odd because only 7 x 1 equals seveb

answer: h=5.6 i don't really know how to put into words, so i though i would let you do that. i hope i was able to some

step-by-step explanation:

18 = 6(h)+1/2

18 = 6h+1/2

18-1 = 6h/2

17 = 3h

17/3 = h

h = 5.6