help me to solve this
2007-07-05 14:53:12 · 9 answers · asked by NOMEGA 1 in Science & Mathematics ➔ Mathematics
is f(x^2 - 5)=[(x - 2)^2-5] or (x^2 - 5) - 2?
2007-07-05 14:58:59 · update #1
By substitution you get:
f(x^2 - 5) = (x^2 - 5) - 2 = x^2 - 7, and
g(x + 1) = (x + 1)^2 - 5 = x^2 + 2x - 4.
Therefore:
f(x^2 - 5) - 3g(x + 1) = - 2x^2 - 6x + 5.
2007-07-05 15:38:04 · answer #1 · answered by fernando_007 6 · 2⤊ 0⤋
f(x^2 – 5) – 3g(x + 1)
= x^2 - 5 – 2 – 3[(x + 1)^2 – 5]
= x^2 - 5 – 2 – 3[x^2 + 2x + 1 – 5]
= x^2 – 7 – 3(x^2 + 2x – 4)
= x^2 – 7 – 3x^2 – 6x + 12
= –2x^2 – 6x + 5
2007-07-10 23:33:58 · answer #2 · answered by semyaza2007 3 · 2⤊ 0⤋
f(x^2 - 5) - 3g(x+1)
by subsitution
x^2 -5 -2 - 3((x+1)^2 - 5)
x^2 - 7 - 3(x^2 + 2x +1 - 5)
x^2 - 7 - 3(x^2 +2x - 4)
x^2 - 7 - 3x^2 -6x + 12
-2x^2 -6x + 5
2007-07-12 23:33:09 · answer #3 · answered by trader 4 · 1⤊ 0⤋
I think melislyn is correct
f(x) = x-2
g(x) = x^2-5
f(x^2-5 - 3g(x+1)
= [(x-2)^2 - 5] - 3[x^2-5+1]
=[(x-2)(x-2) - 5] - 3x^2 + 15 - 3
=x^2 - 4x + 4 - 5 - 3x^2 + 15 - 3
= -2x^2 - 4x + 11
2007-07-10 21:53:05 · answer #4 · answered by ferdie 2 · 0⤊ 1⤋
Here's what I got:
f((x-2)^2 -5) - 3((x^2-5)+1)
= (x-2)^2 + -5 + -3(x^2 + -5) + (-3*1) multiply through
= (x-2)(x-2) + -5 + -3x^2 + 15 + -3 multiply and simplifly
= x^2 -2x -2x + -1 + -3x^2 + 12 reorder to clarify
= x^2 + -3x^2 + -4x + -1 + 12 simplify
= -2x^2 + -4x + 11
2007-07-05 22:53:12 · answer #5 · answered by melislyn 5 · 0⤊ 1⤋
f(x^2 - 5) - 3g(x + 1)
= (x^2 - 5)+2 - 3[(x + 1)^2 - 5]
= -2x^2 - 6x + 9
2007-07-05 21:57:15 · answer #6 · answered by sahsjing 7 · 0⤊ 1⤋
f(x) = x - 2
f (x² - 5) = (x² - 5) - 2 = x² - 7 = A
g(x) = x² - 5
g(x + 1) = (x + 1)² - 5 = x² + 2x - 4 = B
A - 3 B = (x² - 7) - 3.(x² + 2x - 4)
A - 3 B = - 2x² - 6x + 5
2007-07-09 13:42:39 · answer #7 · answered by Como 7 · 2⤊ 0⤋
como and fernando are correct
-2x^2 - 6x + 5
shsjing is off by 4 cause 2 was added instead of subtracted (net change of 4)
2007-07-10 14:07:11 · answer #8 · answered by Shawn D 2 · 1⤊ 1⤋
lol i see 3 different answers
2007-07-07 19:48:31 · answer #9 · answered by ۞_ʞɾ_ 6 · 1⤊ 2⤋