Say I have the problem solved up to this point: 16x^10+28x^5+28x^5+59.
Do I add the powers of 5 or leave alone and just add the numbers?
2007-07-07 14:42:11 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Add the numbers.
16x^10+28x^5+28x^5+59
16x^10+56x^5+59
2007-07-07 14:45:54 · answer #1 · answered by Runa 7 · 0⤊ 0⤋
The two terms of x^5 - both 28x^5 - are *common terms* and can be added together by adding their coefficients (the 28s) but keeping the same exponent.
So your simplification of that would be:
16x^10 + 56x^5 + 59
2007-07-07 21:48:21 · answer #2 · answered by Tony The Dad 3 · 1⤊ 0⤋
Just add the numbers. 16x^10 + 56x^5 + 59
2007-07-07 21:48:09 · answer #3 · answered by Kyle 4 · 1⤊ 0⤋
16x^10+56x^5+59
2007-07-07 21:50:43 · answer #4 · answered by john 4 · 1⤊ 0⤋
Consider the following expression:-
3x² + 4x + 5x² + 3x - 2x²
You now group "things that are the same":-
(3x² + 5x² - 2x²) + (4x + 3x)
to give
6x² + 7x
In your example the things that are the same are the terms x^5 to give:-
16x^10 + 56x^5 + 59
2007-07-11 15:44:59 · answer #5 · answered by Como 7 · 0⤊ 0⤋
Just add the items in the same group, i. e.:
16x^10 + 56^5 + 59
You need not tire your thought after that.
2007-07-07 21:55:50 · answer #6 · answered by Jun Agruda 7 · 3⤊ 0⤋