Use Pythagorean theorem:
9i-j = sqrt (9^2 - 1^2) = sqrt(81-1) = sqrt80
now divide both terms in V by that:
u = 9/sqrt(80)i - 1/sqrt(80)j
see attached picture:
Answer:
89,000 I think
Step-by-step explanation:
just subtract 125,000 by 36,000
although Its a guess
Answer:
The correct option is A. x – 1 < n < 3x + 5
Step-by-step explanation:
In a triangle sum of any two sides is always greater than the third side.
Now, the sides of the triangle are given to be :
2x + 2, x + 3 , n
Now, first take 2x + 2 and x + 3 as two sides and the side of length n as third side.
By using the property that sum of two sides is always greater than the third side in a triangle.
⇒ 2x + 2 + x + 3 > n
⇒ 3x + 5 > n ......(1)
Now, take n and x + 3 as two sides and the side of length 2x + 2 as the third side of triangle.
So, by the property, we have :
n + x + 3 > 2x + 2
⇒ n > x - 1 ...........(2)
From both the equations (1) and (2) , We get :
x – 1 < n < 3x + 5
Therefore, The correct option is A. x – 1 < n < 3x + 5
(√3 - <em>i </em>) / (√3 + <em>i</em> ) × (√3 - <em>i</em> ) / (√3 - <em>i</em> ) = (√3 - <em>i</em> )² / ((√3)² - <em>i</em> ²)
… = ((√3)² - 2√3 <em>i</em> + <em>i</em> ²) / (3 - <em>i</em> ²)
… = (3 - 2√3 <em>i</em> - 1) / (3 - (-1))
… = (2 - 2√3 <em>i</em> ) / 4
… = 1/2 - √3/2 <em>i</em>
… = √((1/2)² + (-√3/2)²) exp(<em>i</em> arctan((-√3/2)/(1/2))
… = exp(<em>i</em> arctan(-√3))
… = exp(-<em>i</em> arctan(√3))
… = exp(-<em>iπ</em>/3)
By DeMoivre's theorem,
[(√3 - <em>i </em>) / (√3 + <em>i</em> )]⁶ = exp(-6<em>iπ</em>/3) = exp(-2<em>iπ</em>) = 1
