math geniuses



pls help with this question!






and pls look at my other questions!!! tysm
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math geniuses pls help with this question and pls look at my other questions tysm class=

Respuesta :

Problem 1, part (a)

[tex]||v|| = \sqrt{(-9)^2+(7)^2} = \sqrt{130}\\\\||w|| = \sqrt{(7)^2+(5)^2} = \sqrt{74}\\\\\cos(\theta) = \frac{v \cdot w}{||v||*||w||}\\\\\cos(\theta) = \frac{(-9)*(7)+7*5}{\sqrt{130}*\sqrt{74}}\\\\\cos(\theta) \approx \frac{-28}{98.081599}\\\\\cos(\theta) \approx -0.285477\\\\\theta \approx \cos^{-1}(-0.285477)\\\\\theta \approx 106.587363\\\\\theta \approx 107\\\\[/tex]

Answer: 107 degrees

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Problem 1, part (b)

The previous result (107) is neither 0, nor 90, nor 180. This means the vectors are neither parallel nor perpendicular.

Answer: Neither

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Problem 2, part (a)

We follow the same idea as problem 1, part (a). Just with different numbers of course.

[tex]||u|| = \sqrt{(-3)^2+(2)^2} = \sqrt{13}\\\\||r|| = \sqrt{(-6)^2+(4)^2} = \sqrt{52}\\\\\cos(\theta) = \frac{u \cdot r}{||u||*||r||}\\\\\cos(\theta) = \frac{(-3)*(-6)+2*4}{\sqrt{13}*\sqrt{52}}\\\\\cos(\theta) = \frac{26}{26}\\\\\cos(\theta) = 1\\\\\theta = \cos^{-1}(1)\\\\\theta = 0[/tex]

Answer:  0

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Problem 2, part (b)

The lines are parallel because of the angle of 0 degrees between the vectors. The angles point in the same direction.

Answer: Parallel