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Hitman42 [59]
3 years ago
14

HELP FOR BRAINLESS !!!!

Mathematics
2 answers:
garik1379 [7]3 years ago
8 0

Answer:

the blue one

Step-by-step explanation:

Kruka [31]3 years ago
5 0

Answer:

ok

Step-by-step explanation:

its the first one

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Find the angle between vector u=2i+sqrt(11)j and v=-3i-2j to the nearest degree
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Step-by-step explanation:

We have the components of the vectors u and v.

Then, to find the angle between them, perform the following steps:

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Then, the product scalar u * v is:

u * v = (2)(-3) + (\sqrt{11})(-2)

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2) Calculation of the magnitude of both vectors.

| u | = \sqrt{(2) ^ 2 + (\sqrt{11})^2}\\\\| u | = \sqrt{4+11}\\\\| u | = \sqrt{15}

| v | = \sqrt{(- 3) ^ 2 + (- 2) ^ 2}\\\\| v | = \sqrt{9 +4}\\\\| v | = \sqrt{13}

3) Now that you know the product point between the two vectors and the magnitude of each, then use the following formula to find an angle

u * v = | u || v |cos(\alpha)

-12.633 = \sqrt{15}\sqrt{13}*cos(\alpha)\\\\cos(\alpha) = \frac{-12.633}{\sqrt{15}\sqrt{13}}\\\\arcos(\frac{-12.633}{\sqrt{15}\sqrt{13}}) = \alpha\\\\\alpha =155\°

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