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kipiarov [429]
3 years ago
5

What is the magnitude of the position vector whose terminal point is (6, -4)?

Mathematics
1 answer:
Marysya12 [62]3 years ago
5 0

Answer:

  2√13

Step-by-step explanation:

The distance formula is useful for this. One end of the vector is (0, 0), so the measure of its length is ...

  d = √((x2 -x1)² +(y2 -y1)²) = √((6 -0)² +(-4-0)²)

  = √(36 +16) = √52 = √(4·13)

  d = 2√13 = |(6, -4)|

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AABC ~ADEF<br> F<br> 21<br> D<br> B<br> N<br> 28<br> E<br> C<br> Solve for N.<br> 28 = ☆<br> N=[?]
sergij07 [2.7K]

Answer:

N = 3

Step-by-step explanation:

\frac{28}{4}  =  \frac{21}{N}  \\  \\ 28N = 21 \times 4 \\ 28N = 84 \\  \\ N =  \frac{84}{28}  \\  \\ N = 3

I hope I helped you^_^

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Step-by-step explanation:

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Akimi4 [234]

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Ad libitum [116K]

Answer:

The answer to your question is:

Step-by-step explanation:

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c.- Find the fifth term = 15u⁴v¹²

4 0
3 years ago
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