(2^1-√3)^1+√3=?
How do you get the conclusion?
2007-03-26 23:17:31 · 11 answers · asked by thatfella 1 in Science & Mathematics ➔ Mathematics
Anything to the first power is just itself...
Ex: 2^1 = 2
so
(2^1 - sqrt 3)^1 + sqrt 3
= 2 - sqrt 3 + sqrt 3
= 2
2007-03-26 23:20:07 · answer #1 · answered by Mathematica 7 · 0⤊ 0⤋
= 2 - â3 + â3
= 2
2007-03-27 06:31:54 · answer #2 · answered by Como 7 · 0⤊ 0⤋
(2^1 - â3)^1 + â3
= 2 - â3 + â3
= 2
2007-03-27 06:25:54 · answer #3 · answered by rooster1981 4 · 0⤊ 0⤋
Any number to the power of 1 is the same number.
So, 2^1 = 2.
2 - â3 + â3 = 2
2007-03-27 06:31:42 · answer #4 · answered by Norrie 7 · 0⤊ 0⤋
(2^1-â3)^1+â3
= (2^1-â3) +â3
= 2^1-â3+â3
But,
-3 + 3 = 3 - 3
= 0
Therefore,
2^1 - 0
= 2^1
= 2
2007-03-27 06:27:00 · answer #5 · answered by Nirmal87 2 · 0⤊ 0⤋
I'm taking a stab in the dark here but is your question:
(2^(1-sqrt3))^(1+sqrt3) by any chance?
If so, use your power rules to say that this equals
2^((1-sqrt3)(1+sqrt3))
=
2^(1^2 - (sqrt3)^2) ------------- difference of two squares
=
2^(1-3)
=
2^-2
=
1/4
2007-03-27 06:31:26 · answer #6 · answered by SteveK 5 · 0⤊ 0⤋
x^1 = x so
2^1=2
(2-â3)^1 = 2-â3
so 2 - â3 + â3 = 2
2007-03-27 06:31:08 · answer #7 · answered by Anonymous · 0⤊ 0⤋
(2^(1- (3^.5))) ^(1+(3^.5))
Is the same as
2 to the power of (1- (3^.5) ) x (1+(3^.5))
(1-(3^.5))x(1+(3^.5))=
1- (3^.5) + (3^.5) - 3 = -2
2 ^(-2) = (1/2)^2 = 0,25
2007-03-27 06:43:34 · answer #8 · answered by Anonymous · 0⤊ 0⤋
2^1 - sqrt3^1 + sqrt3 = 2 - sqrt3 + sqrt3 = 2
2007-03-27 06:27:29 · answer #9 · answered by bach 2 · 0⤊ 0⤋
anynumber to its power is always equal to that number.
So (2^1-sqroot3)^1+sqroot3=2^1-sqroot3+sqroot3
=2^1 (cancelling + and - sqroot3)=2
2007-03-31 04:31:31 · answer #10 · answered by sandy 2 · 0⤊ 0⤋