What is the sum of the digits of all
2-digit numbers from 10 to 99?

(99)(100)/2 - (9)(10)/2 = 4905

Fast way to solve: allow leading zero's temporarily, then first digit can be any of 0..9 and so can second digit, so each digit appears 10 times in each position, so sum is 20 * (9)(10)/2, except that we must subtract sum of 1..9, so it's 19 * (9)(10)/2 = 855.

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: