can this be done this way ? I find its time consuming to expand the powers, add and then covert it to power agian. 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 = 2^2 + 2^2+2^3+2^4+2^5+2^6+2^7+2^8 = 2^2 [ 1+1+2+2^2+2^3+2^4+2^5+2^6] =2^2 . 2^2 [1+1+2+2^2+2^3+2^4] = 2^2 . 2^2 .2^2 [1+1+2+2^2 ] =2^2 . 2^2 .2^2 .2^2 ...

If xy+z = x(y+z), which of the following must be true? 1. x=0 and z=0 2. x=1 and y=1 3. y=1 and z=0 4. x=1 or y=0 5. x=1 or z=0 I was able to solve for x as xy+z = x(y+z) xy+z = xy +xz xz = z x =1 I randomly marked option 2 and moved on as i dint know how to evaluate z and y. Please Help. OA is 5.

Oops ....realized the mistake
xz = z

xz-z = 0
z(x-1)=0
x =1, z=0

The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers than T? 1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T. 2) The median of the integers in S is greater than the median of the integers ...

Infact Ron, OA is A

i have the screen shot...but i am not able to post it :(