2006-09-07 12:26:03 · 5 answers · asked by kayteedella 1 in Science & Mathematics ➔ Mathematics
As has already been said, (169-9x^2) = (13-3x)*(13+3x). Now determine when each one of these equals zero: 13-3x=0 when x=13/3; 13+3x=0 when x=-13/3. Now time to analyze: when x is greater then 13/3, (13-3x) is less then zero and (13+3x) is greater then zero; therefore, their product is negative. Therefore the domain must not be greater then 13/3. When x<-13/3, (13-3x) will be positive and (13+3x) will be negative, their product would then be negative; therefore the domain can not be less then -13/3. Answer: -13/3<=x<=13/3.
2006-09-07 12:42:27 · answer #1 · answered by bruinfan 7 · 0⤊ 0⤋
sqrt(169 - 9x^2) >= 0
169 - 9x^2 >= 0
-9x^2 >= -169
x^2 <= (169/9)
x <= (-13/3) , x >= (13/3)
Domain :
(-13/3) <= x <= (13/3)
2006-09-07 20:41:46 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
(169 - 9x² ) >= 0
(13+3x) (13-3x) >=0
13+ 3x >= 0
3x >= -13
OR
13 - 3x >= 0
-3x >= -13
x <= 13/3
Therfore -13/3 <= x <= 13/3
Cheers.
2006-09-11 00:59:52 · answer #3 · answered by isz_rossi 3 · 0⤊ 0⤋
(169 - 9x^2)
=(13-3x)(13+3x)
3x = -13 ; 3x =13
Therefore X = -13/3 or 13/3
2006-09-08 07:08:42 · answer #4 · answered by Forgettable 5 · 0⤊ 0⤋
This is a difference between two squares so the factors are (13-3x) and (13+3x)
2006-09-07 19:28:29 · answer #5 · answered by Nelson_DeVon 7 · 0⤊ 0⤋