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bagirrra123 [75]
3 years ago
15

Find the distance between v = [1; 2; 3]t and w = [-2; 1;2]t

Mathematics
1 answer:
saveliy_v [14]3 years ago
8 0

Answer:

  |w-v| = (√11)t

Step-by-step explanation:

Using the distance formula, you find it to be (for t=1) ...

  d = √((-2-1)² +(1-2)² +(2-3)²) = √(9 +1 +1) = √11

The distance will be multiplied by t, so is ...

  |w-v| = (√11)t

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Solve each quadratic equation by completing the square. 6. x2 + 2x = 8 7. x2 - 6x = 16 8. x2 - 18x = 19 9. x2 + 3x = 3 10. x2 +
Andrei [34K]
Lets get started :)

These questions have asked us to solve by completing the square.
How do we? I have attached a picture, which will explain

6. x² + 2x = 8
→ b is the coefficient of x, which is 2
→ We take half of 2 and square it. Then, we add it to either side

x² + 2x + (\frac{2}{2} )^2 = 8 + ( \frac{2}{2})^2
x² + 2x + 1 = 8 + 1
( x + 1 )( x + 1 ) = 9
( x + 1 )² = 9
\sqrt{( x + 1 )^2} = \sqrt{9}
x + 1 = + 3 or x + 1 = - 3
    x = 2     or     x = - 4

7. x² - 6x = 16

→ We do the same thing we did in the previous question

x² - 6x + (\frac{6}{2})^2 = 16 +  (\frac{6}{2})^2 
x² - 6x + 9 = 16 + 9
(x - 3)² = 25
\sqrt{(x-3)^2} =  \sqrt{25}
x - 3 = + 5 or x - 3 = - 5
   x = 8      or       x = - 2

8. x² - 18x = 19

x² - 18x + ( \frac{18}{2} )^2 = 19 + ( \frac{18}{2})^2
x² - 18x + 81 = 19 + 81
( x - 9 )( x - 9 ) = 100
( x - 9 )² = 100
\sqrt{(x-9)^2} =  \sqrt{100}
x - 9 = + 10 or x - 9 = -10
   x = 19      or      x = - 1

9. x² + 3x = 3

x² + 3x + ( \frac{3}{2} )^2 = 3 +  (\frac{3}{2} )
x^2 + 3x +  \frac{9}{4} = 3 +  \frac{9}{4}
x^{2} + 3x +  \frac{9}{4} =  \frac{21}{4}
(x^2 +  \frac{3}{2} ) ( x^2 +  \frac{3}{2} ) =  \frac{21}{4}
( x^2 + \frac{3}{2} )^2 =  \frac{21}{4}
\sqrt{ x^2 + \frac{3}{2} } =  \sqrt{ \frac{21}{4} } 
x +  \frac{3}{2} = + \frac{ \sqrt{21} }{2} or x +\frac{3}{2} = -  \frac{ \sqrt{21} }{2}
x = \frac{-3+ \sqrt{21} }{2} or x = \frac{-3 - \sqrt{21}}{2}


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