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Nutka1998 [239]
3 years ago
9

What is the solution to the linear equation? 4b+6=2-b+4

Mathematics
2 answers:
Morgarella [4.7K]3 years ago
7 0
To start we need to put all the b's on the same side.

4b + 6 = 2 - b + 4
To move the b on the right side to the left side, add b to both sides.
4b + 6 + b = 2 - b + 4 + b

Simplify. 4b + b = 5b and -b + b = 0 so...
The equation is now:  5b + 6 = 2 + 4

Now move 6 to the right side so we can isolate b.
5b + 6 - 6 = 2 + 4 - 6
Simplify.
5b = 0
Solve for b.
b = 0/5
b = 0


vaieri [72.5K]3 years ago
3 0
4b+6= 2 - b + 4    (combine terms: 2 + 4 = 6)
4b + 6 = 6 - b  (add the b to the other side to get rid of the negative)
5b + 6 = 6   (subtract 6 from one side to the other)
5b = 0  (divide by 5 on both sides)
b = 0  

plug it back in




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w + g = 62        (w = white and g = green marbles).

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De moirve's <br> (√3-i ÷ √3+i)^6 = 1
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(√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

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