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vampirchik [111]
3 years ago
15

Let v be a vector with initial point (–5, –11) and terminal point (6, 2). What is Norm of vector v?

Mathematics
2 answers:
GuDViN [60]3 years ago
4 0

Answer:

B. \sqrt{290}

Step-by-step explanation:

On Edge, just finished the test.

Anvisha [2.4K]3 years ago
4 0

Answer:

B maybe?

Step-by-step explanation:

Good luck in the rest of the year

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Below

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56% of her eggs were medium in may

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3 years ago
Trigonometry:<br> Simplify the expression.<br><br> (cot²α - 4) / (cot²α - cotα - 6)
Anettt [7]
<span>(cot²α - 4) / (cot²α - cotα - 6);  Let x = </span><span>cotα

Therefore  </span><span>(cot²α - 4) / (cot²α - cotα - 6)   = </span><span>(x²- 4) / (x² - x - 6)
</span>
(x²- 4) =  <span>x² - </span><span>2² = (x-2)(x+2)  . Difference of two squares.

</span><span>(x² - x - 6);  This is a quadratic expression,
Multiply the first and last terms = -6</span>x²
we think of two expression that multiply to give -6x² and add up to give -x (Middle term)

Those expression are 2x and -3x
(x² - x - 6) = <span>(x² +2x-3x - 6) = x(x+2) -3(x+2) = </span>(x+2)<span><span>(x-3)

</span> Recall </span><span>(x²- 4) / (x² - x - 6) = (</span><span>(x-2)(x+2)) / (</span><span>(x+2)<span>(x-3)) = </span></span><span>(x-2)/</span><span><span>(x-3). Cancelling out.</span> </span>

Recall  x = <span>cotα, therefore:</span>
(x-2)/<span>(x-3) = </span>(cotα-2)/(cotα-3)

(cot²α - 4) / (cot²α - cotα - 6)  =  (cotα-2)/(cotα-3)

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