
William S. answered 10/16/13
Tutor
4.4
(10)
Experienced scientist, mathematician and instructor - William
1a. f+g = abs (x) + abs (x-3)
1b. f-g = abs (x) - abs (x-3)
1c. f*g = abs [x(x-3)]
1d. f/g = abs[x/(x-3)]
1e. g/f = abs [(x-3)/x]
2a. f+g = {[1/(x+1)] + [2/(x-2)] +1}
2b. f-g = {[1/(x+1)] -[2/(x-2)] -1}
2c. f*g = {x/[(x+1)(x-2)]}
2d. f/g = {(x-2)/[x(x+1)]}
2e. g/f = {[x(x+1)]/(x-2)}
3a. f+g = x2 + [1/(x½)]3b.
3b f-g = x2 - [1/(x½)]
3c. f*g = x3/2
3d. f/g = x5/2
3e. = 1(x5/2)

William S.
Paige,
1a. abs(x)+abs(x-3) = abs(x)+abs(x-3) No further simplification is possible.
1b. abs(x)-abs(x-3) = abs(x)-abs(x-3) No further simplification is possible.
1c. abs(x)*abs(x-3) = abs [x(x-3)] No further simplification is possible.
1d. abs(x)/abs(x-3) = abs[x/(x-3)] No further simplification is possible.
1e. abs(x-3)/abs(x) = abs [(x-3)/x] No further simplification is possible.
2a. [1/(x+1)] + [x/(x-2)] = {[1/(x+1)] + [2/(x-2)] +1} No further simplification is possible.
2b. [1/(x+1)] - [x/(x-2)] = {[1/(x+1)] - [2/(x-2)] -1} No further simplification is possible.
2c. [1/(x+1)] * [x/(x-2)] = {x/[(x+1)(x-2)]} No further simplification is possible.
2d. [1/(x+1)] / [x/(x-2)] = {(x-2)/[x(x+1)]} No further simplification is possible.
2e. [x/(x-2)] / [1/(x+1)] = {[x(x+1)]/(x-2)} No further simplification is possible
3a. (x^2) + (1/√x) = (x^2) + (1/√x)
3b. (x^2) - (1/√x) = (x^2) - (1/√x)
3c. (x^2) * (1/√x) = x3/2 or [(x)*(√x)]
3d. (x^2) / (1/√x) = x5/2 or [(x^2)*(√x)]
3e. (1/√x)/(x^2) = (1/x5/2) or
x-5/2
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10/21/13
Paige N.
Thank you very much Sir! :) You have helped me a lot!
Report
10/21/13

William S.
Paige, you are very welcome. I live to help young people understand mathematics, science, history, English and many other things. You are the only person I've met here who thanked me, and I appreciate it very much.
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10/22/13
Paige N.
10/21/13